21.12 - Logarithmic Functions

In the previous section, we completed a discussion on exponential functions. In this section, we will see logarithmic functions.

Some basic details about logarithmic functions can be written in 3 steps:
1. Consider the exponential function that we saw in the previous section: y = f(x) = b^x.
♦ where b > 0 and b ≠ 1.
2. Here "b" is a constant. x and y are variables.
• If we give an input x, we will get an output y.
• For example, if b = 2 and x = 3, we get: y = 2³ = 8.
• In such a situation, we say that:
Logarithm of 8 to the base 2 is 3
• In general, if y = b^x, we say that:
Logarithm of y to the base b is x.
3. Logarithm of y to the base b is denoted as log_b y.
• So if y = b^x, we can write: log_b y = x.

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Let us see some solved examples:

Solved example 21.41
Evaluate the following logarithms:
(a) log₁₀ 1000 (b) log₄ 16 (c) log₅ 625 (d) log_1/6 36 (e) $\log_9 \frac{1}{531441}$ (f) $\log_{\frac{3}{2}} \frac{27}{8}$
Solution:
Part (a)
1. Let log₁₀ 1000 = x
2. Then we can write: 1000 = 10^x.
3. We have: 1000 = 10³.
4. From (2) and (3), we get: 10^x = 10³.
• In the above equation, bases are the same. So equating the powers, we get: x = 3

Part (b)
1. Let log₄ 16 = x
2. Then we can write: 16 = 4^x.
3. We have: 16 = 4².
4. From (2) and (3), we get: 4^x = 4².
• In the above equation, bases are the same. So equating the powers, we get: x = 2

Part (c)
1. Let log₅ 625 = x
2. Then we can write: 625 = 5^x.
3. We have: 625 = 5⁴.
4. From (2) and (3), we get: 5^x = 5⁴.
• In the above equation, bases are the same. So equating the powers, we get: x = 4

Part (d)
1. Let $\log_{\frac{1}{6}} 36$ = x
2. Then we can write: $36 = \left(\frac{1}{6} \right)^x$.
⇒ $36 = 6^{-x}$
3. We have: 36 = 6².
4. From (2) and (3), we get: 6^−x = 6².
• In the above equation, bases are the same. So equating the powers, we get: x = −2

Part (e)
1. Let $\log_9 \frac{1}{531441}$ = x
2. Then we can write: $\frac{1}{531441}$ = 9^x.
⇒ (531441)⁻¹ = 9^x.
3. We have: 531441 = 9⁶.
⇒ (531441)⁻¹ = (9⁶)⁻¹ = 9⁻⁶.
4. From (2) and (3), we get: 9^x = 9⁻⁶.
• In the above equation, bases are the same. So equating the powers, we get: x = −6

Part (f)
1. Let $\log_{\frac{3}{2}} \frac{27}{8}$ = x
2. Then we can write: $\frac{27}{8} = \left(\frac{3}{2} \right)^x$.
3. We have: $\frac{27}{8} = \left(\frac{3}{2} \right)^3$.
4. From (2) and (3), we get: $\left(\frac{3}{2} \right)^x = \left(\frac{3}{2} \right)^3 $.
• In the above equation, bases are the same. So equating the powers, we get: x = 3

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Now we will see how logarithm can be used as a function. It can be written in 4 steps:
1. We know that, if y = b^x, then: log_b y = x.
2. Consider the expression log_b y = x.
• Here, the input y is being processed to obtain y.
• In other words, y is being subjected to a process. The process is nothing but "finding the logarithm of y". (The solved examples that we saw just above, show us how to find the logarithm of a given number). The resulting logarithm is the output.
3. But for functions,
    ♦ Input values are denoted as x.
      ✰ They are plotted along the x-axis.
    ♦ Output values are denoted as y.
      ✰ They are plotted along the y-axis
• So in the expression log_b y = x, we need to interchange x and y. We get: log_b x = y.
4. So we get a function in which, input x is processed to give output y.
• The process is nothing but "finding the logarithm of x". The resulting logarithm is the output y.
• We can write: y = f(x) = log_b x.
This is called logarithmic function.

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Let us write the important features about logarithmic functions. It can be written in 7 steps:
1. The general form of the logarithmic function is: f(x) = log_b x.
   ♦ b should be greater than zero.
   ♦ b should not be equal to 1.

2. Let us see why b should be greater than zero. It can be written in (iii) steps.
(i) Suppose that, b = −2.
Then the function will be y = f(x) = log₍₋₂₎ x
(ii) While plotting the graph, when the input is 2, we need to find log₍₋₂₎ 2
• Let us try:
   ♦ Assume log₍₋₂₎ 2 = y
   ♦ Then we can write: 2 = (−2)^y.
   ♦ ⇒ −2 = 2^(1/y).
   ♦ This is impossible because, no power of 2 will give a negative value.
(iii) To avoid such situations, we avoid −ve numbers altogether.

3. Let us see why b should not be equal to one. It can be written in (iii) steps.
(i) Suppose that, b = 1.
Then the function will be y = f(x) = log₁ x
(ii) While plotting the graph, when the input is 2, we need to find log₁ 2
• Let us try:
   ♦ Assume log₁ 2 = y
   ♦ Then we can write: 2 = (1)^y.
   ♦ This is impossible because, all powers of 1 give 1.
(iii) To avoid such a situation, we avoid 1.

4. Fig.21.18 below shows the graphs of some simple logarithmic functions.

Fig.21.18

• Red, and yellow belong to the category: b > 1
• Green and magenta belong to the category: 0 < b < 1
• White shows b = e. It belong to the category: b > 1

5. From the graphs, we see that,
• When b >1:
   ♦ As x increases towards ∞, f(x) also approaches ∞.
         ✰ Red, yellow and white are rising up.
   ♦ As x decreases towards zero, f(x) approaches −∞.
         ✰ Red, yellow and white are falling down.
• When 0 < b < 1:
    ♦ As x increases towards ∞, f(x) approaches −∞.
          ✰ Green and magenta are falling down.
    ♦ As x decreases towards zero, f(x) approaches ∞.
          ✰ Green and magenta are rising up.


6. From the graphs, we see that:
• Input x cannot be a −ve number.
The reason can be demonstrated in (iii) steps:
(i) Suppose that, x = −1000 and b = 10
(Recall that, b cannot be −ve)
So we want to find y, where y = log₁₀ (−1000)
(ii) Let us try:
   ♦ We can write: −1000 = (10)^y.
   ♦ This is impossible because, no power of 10 will give a negative value.
(iii) So for logarithmic functions, input can never be −ve.
We will see the actual proof in higher classes.

• Since no input can be −ve, we say that:
Domain of the logarithmic function is R⁺.

7. From the graphs, we get the following information also:
• The output of a log function can never be zero.
   ♦ Note that, none of the graphs touch the y-axis.
• The range of a log function is (−∞,∞).
• For a log function, the point (1,0) will be always available.
   ♦ This is because, any number raised to the power zero, is 1. Which implies: Whatever be the base, logarithm of 1 is zero.

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Now we will discuss the fact that, log function is the inverse of the exponential function. It can be written in 4 steps:
1. Fig.21.19 shows the graphs of three functions:
(i) y = f(x) = e^x. (red color)
(ii) y = f(x) = x. (magenta color)
(iii) y = f(x) = log_e x. (yellow color)

Fig.21.19

2. Mark any point P on the magenta line.
• Through P, draw a white dashed line perpendicular to the magenta line.
• The white dashed line,
   ♦ intersects the red curve at Q.
   ♦ intersects the yellow curve at R.
• The distances PQ and PR are equal.

3. PQ and PR are equal because,
   ♦ red and yellow curves are mirror images of each other.
   ♦ magenta line is the mirror line.

4. y = e^x is the inverse of y = log_e x and vice versa.
If we combine them as a composite function, we will get an identity function. This can be demonstrated in (iii) steps:
(i) Let f(x) = e^x and g(x) = log_e x.
• Then f(g(x)) = f(log_e x) = $\rm{e^{\log_e x}}$
• Let $\rm{e^{\log_e x}}$ = u
⇒ log_e u = log_e x
⇒ u = x
⇒ f(g(x)) = u = x
(ii) Similarly, g(f(x)) = g(e^x) = log_e (e^x).
• Let log_e (e^x) = v
⇒ e^x = e^v.
⇒ v = x
⇒ g(f(x)) = v = x
(iii) From (i) and (ii), we get: f(g(x)) = g(f(x)) = x
• That means, f(x) is the inverse of g(x) and vice versa.
• In other words, y = e^x is the inverse of y = log_e x and vice versa.
• This true for all acceptable values of base b. We can write:
   ♦ y = 10^x is the inverse of y = log₁₀ x and vice versa.
   ♦ y = 2^x is the inverse of y = log₂ x and vice versa.
   ♦ y = e^x is the inverse of y = log_e x and vice versa.
   ♦ etc.,

Based on this information, let us see a solved example:

Solved example 21.42
Is it true that $\rm{x = e^{\log x}}$ for all real x?
Solution:
1. Consider $\rm{x = e^{\log x}}$.
• It is a composite of two functions:
    ♦ $\rm{y = e^x}$
    ♦ $\rm{y = \log_e x}$
(Recall that, in this chapter, we write $\rm{\log_e x}$ simply as $\rm{\log x}$)
2. Each of the two functions is inverse of the other.
• So whatever is the input, the output will be same as input.
3. Now suppose that, the input x is a −ve number.
• Then the "$\rm{\log x}$" portion of the composite function will not be able to process the input x. This is because, logarithm of −ve numbers does not exist.
4. Therefore, the equation $\rm{x = e^{\log x}}$ is true only when input x is +ve.

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In a logarithm function, if e is taken as the base, then we denote it as ln. This can be explained in 4 steps:
1. Recall that, in the case of the exponential function y = b^x, if the base is 10, we call it common exponential function.
• In a similar way, in the case of the logarithmic function y = log_b x, if the base is 10, we call it common logarithmic function.
2. Recall that, in the case of the exponential function y = b^x, if the base is e, we call it natural exponential function.
• In a similar way, in the case of the logarithmic function y = log_b x, if the base is e, we call it natural logarithmic function.
3. So the natural logarithmic function is: y = log_e x
• It can be written in a short form as: y = ln x.
4. In this chapter, if the base is not specified, it means natural logarithm.
• So, in this chapter, if we see y = log 5, then it means, y = the natural logarithm of 5.

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We have completed a basic discussion on logarithmic functions. In the next section, we will see properties of logarithmic functions.

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17.8 - Properties of Invertible Functions

In the previous section, we saw two solved examples related to invertible functions. We saw composite of composite functions also. In this section, we will see two more solved examples. We will also see a new property of invertible functions.

Solved example 17.27
Consider two functions:

f: {1,2,3}→{a,b,c} defined as:
f(1) = a, f(2) = b, f(3) = c

g: {a,b,c}→{apple, ball, cat} defined as:
g(a) = apple, g(b) = ball, g(c) = cat.

I. Show that, f, g and (g∘f) are invertible.
II. Find out f^-1, g^-1 and (g∘f)^-1.
III. Show that (g∘f)^-1 = f^-1∘g^-1

Solution:
Part (I): Proving that f, g and (g∘f) are invertible.
1. f is a function. Domain is {1,2,3} and codomain is {a,b,c}.
• We know that f is a set. The elements of this set are ordered pairs of the form (x,y).
• Based on the given data, we can write all the ordered pairs in f:
f = {(1,a), (2,b), (3,c)}
• We see that:
   ♦ Each element in the domain has a unique image. So f is one-one.
   ♦ All elements in the codomain are images. So f is onto.
• Since f is both one-one and onto, it is invertible.

2. g is a function. Domain is {a,b,c} and codomain is {apple, ball, cat}.
• We know that g is a set. The elements of this set are ordered pairs of the form (x,y).
• Based on the given data, we can write all the ordered pairs in g:
g = {(a,apple), (b,ball), (c,cat)}
• We see that:
   ♦ Each element in the domain has a unique image. So g is one-one.
   ♦ All elements in the codomain are images. So g is onto.
• Since g is both one-one and onto, it is invertible.

3. (g∘f)(x) = g(f(x))
• For (g∘f),
   ♦ domain is the domain of f which is {1,2,3}
   ♦ codomain is the codomain of g which is {apple, ball, cat}
• So we get:
(g∘f)(1) = g(f(1)) = g(a) = apple
(g∘f)(2) = g(f(2)) = g(b) = ball
(g∘f)(3) = g(f(3)) = g(c) = cat

• We know that (g∘f) is a set. The elements of this set are ordered pairs of the form (x,y).
• Based on the above results, we can write all the ordered pairs in (g∘f):
(g∘f) = {(1,apple), (2,ball), (3,cat)}
• We see that:
   ♦ Each element in the domain has a unique image. So (g∘f) is one-one.
   ♦ All elements in the codomain are images. So (g∘f) is onto.
• Since (g∘f) is both one-one and onto, it is invertible.

Part (II): Finding f^-1, g^-1 and (g∘f)^-1.
1. Finding f^-1:
(i) In part (I), we saw that:
f = {(1,a), (2,b), (3,c)}
• So we can write the reverse of f. We will call it p.
• Since f is one-one and onto, the reverse can be easily written:
p = {(a,1), (b,2), (c,3)}
• The domain of p is {a,b,c}. And codomain is {1,2,3}

(ii) Now we have both f and p. We can calculate (p∘f)
(p∘f) means, output of f is used as the input of p.
• The two functions are:
f: {1,2,3}→{a,b,c}
p: {a,b,c}→{1,2,3}
• We know that, (p∘f) directly connects the domain of f to the codomain of p. So (p∘f) is from {1,2,3} to {1,2,3}.
• When the input is 1, we get:
(p∘f)(1) = p(f(1)) = p(a) = 1
• When the input is 2, we get:
(p∘f)(2) = p(f(2)) = p(b) = 2
• When the input is 3, we get:
(p∘f)(3) = p(f(3)) = p(c) = 3

• So whatever input we give, the output will be the same. That means, (p∘f) is an identity function.
• The inputs are taken from {1,2,3}. So we can write:
(p∘f) = I_{1,2,3}.

(iii) Similarly, we can calculate (f∘p)
(f∘p) means, output of p is used as the input of f.
• The two functions are:
f: {1,2,3}→{a,b,c}
p: {a,b,c}→{1,2,3}
• We know that, (f∘p) directly connects the domain of p to the codomain of f. So (f∘p) is from {a,b,c} to {a,b,c}.
• When the input is a, we get:
(f∘p)(a) = f(p(a)) = f(1) = a
• When the input is 2, we get:
(f∘p)(b) = f(p(b)) = f(2) = b
• When the input is 3, we get:
(f∘p)(c) = f(p(c)) = f(3) = c

• So whatever input we give, the output will be the same. That means, (f∘p) is an identity function.
• The inputs are taken from {a,b,c}. So we can write:
(f∘p) = I_{a,b,c}.

(iv). Let us write a summary:
• We are given a function f: {1,2,3}→{a,b,c}
• We wrote the reverse function p: {a,b,c}→{1,2,3}
• From (ii), we got: (p∘f) = I_{1,2,3}
• From (iii), we got: (f∘p) = I_{a,b,c}
• So f is invertible.
• Also, the inverse of f = f^-1 = p
• That means:
The inverse of f: {1,2,3}→{a,b,c} is:
p: {a,b,c}→{1,2,3}

2. Finding g^-1:
(i) In part (I), we saw that:
g = {(a,apple), (b,ball), (c,cat)}
• So we can write the reverse of g. We will call it q.
• Since g is one-one and onto, the reverse can be easily written:
q = {(apple,a), (ball,b), (cat,c)}
• The domain of q is {apple, ball, cat}. And codomain is {a,b,c}

(ii) Now we have both g and q. We can calculate (q∘g)
(q∘g) means, output of g is used as the input of q.
• The two functions are:
g: {a,b,c}→{apple, ball, cat}
q: {apple, ball, cat}→{a,b,c}
• We know that, (q∘g) directly connects the domain of g to the codomain of q. So (q∘g) is from {a,b,c} to {a,b,c}.
• When the input is a, we get:
(q∘g)(a) = q(g(a)) = q(apple) = a
• When the input is b, we get:
(q∘g)(b) = q(g(b)) = q(ball) = b
• When the input is c, we get:
(q∘g)(c) = q(g(c)) = q(cat) = c

• So whatever input we give, the output will be the same. That means, (q∘g) is an identity function.
• The inputs are taken from {a,b,c}. So we can write:
(q∘g) = I_{a,b,c}.

(iii) Similarly, we can calculate (g∘q)
(g∘q) means, output of q is used as the input of g.
• The two functions are:
g: {a,b,c}→{apple, ball, cat}
q: {apple, ball, cat}→{a,b,c}
• We know that, (g∘q) directly connects the domain of q to the codomain of g. So (g∘q) is from {apple, ball, cat} to {apple, ball, cat}.
• When the input is apple, we get:
(g∘q)(apple) = g(q(apple)) = g(a) = apple
• When the input is 2, we get:
(g∘q)(ball) = g(q(ball)) = g(b) = ball
• When the input is 3, we get:
(g∘q)(cat) = g(q(cat)) = g(c) = cat

• So whatever input we give, the output will be the same. That means, (g∘q) is an identity function.
• The inputs are taken from {apple, ball, cat}. So we can write:
(g∘q) = I_{{apple, ball, cat}}.

(iv). Let us write a summary:
• We are given a function g: {a,b,c}→{apple, ball, cat}
• We wrote the reverse function q: {apple, ball, cat}→{a,b,c}
• From (ii), we got: (q∘g) = I_{a,b,c}
• From (iii), we got: (g∘q) = I_{{apple, ball, cat}}
• So g is invertible.
• Also, the inverse of g = g^-1 = q
• That means:
The inverse of g: {a,b,c}→{apple, ball, cat} is:
q: {apple, ball, cat}→{a,b,c}

3. Finding (g∘f)^-1:
(i) In part (I), we saw that:
(g∘f) = {(1,apple), (2,ball), (3,cat)}
• So we can write the reverse of (g∘f). We will call it r.
• Since (g∘f) is one-one and onto, the reverse can be easily written:
r = {(apple,1), (ball,2), (cat,3)}
• The domain of r is {apple, ball, cat}. And codomain is {1,2,3}

(ii) Now we have both (g∘f) and r. We can calculate (r∘(g∘f))
(r∘(g∘f)) means, output of (g∘f) is used as the input of r.
• The two functions are:
(g∘f): {1,2,3}→{apple, ball, cat}
r: {apple, ball, cat}→{1,2,3}
• We know that, (r∘(g∘f)) directly connects the domain of (g∘f) to the codomain of r. So (r∘(g∘f)) is from {1,2,3} to {1,2,3}.
• When the input is 1, we get:
(r∘(g∘f))(1) = r((g∘f)(1)) = r(apple) = 1
• When the input is 2, we get:
(r∘(g∘f))(2) = r((g∘f)(2)) = r(ball) = 2
• When the input is 3, we get:
(r∘(g∘f))(3) = r((g∘f)(3)) = r(cat) = 3

• So whatever input we give, the output will be the same. That means, (r∘(g∘f)) is an identity function.
• The inputs are taken from {1,2,3}. So we can write:
(r∘(g∘f)) = I_{1,2,3}.

(iii) Similarly, we can calculate ((g∘f)∘r)
((g∘f)∘r) means, output of r is used as the input of (g∘f).
• The two functions are:
(g∘f): {1,2,3}→{apple, ball, cat}
r: {apple, ball, cat}→{1,2,3}
• We know that, ((g∘f)∘r) directly connects the domain of r to the codomain of (g∘f). So ((g∘f)∘r) is from {apple, ball, cat} to {apple, ball, cat}.
• When the input is apple, we get:
((g∘f)∘r)(apple) = (g∘f)(r(apple)) = (g∘f)(1) = apple
• When the input is 2, we get:
((g∘f)∘r)(ball) = (g∘f)(r(ball)) = (g∘f)(2) = ball
• When the input is 3, we get:
((g∘f)∘r)(cat) = (g∘f)(r(cat)) = (g∘f)(3) = cat

• So whatever input we give, the output will be the same. That means, ((g∘f)∘r) is an identity function.
• The inputs are taken from {apple, ball, cat}. So we can write:
((g∘f)∘r) = I_{{apple, ball, cat}}.

(iv). Let us write a summary:
• We are given a function (g∘f): {1,2,3}→{apple, ball, cat}
• We wrote the reverse function r: {apple, ball, cat}→{1,2,3}
• From (ii), we got: (r∘(g∘f)) = I_{1,2,3}
• From (iii), we got: ((g∘f)∘r) = I_{{apple, ball, cat}}
• So (g∘f) is invertible.
• Also, the inverse of (g∘f) = (g∘f)^-1 = r
• That means:
The inverse of (g∘f): {1,2,3}→{apple, ball, cat} is:
r: {apple, ball, cat}→{1,2,3}

Part III: Showing that, (g∘f)^-1 = f^-1∘g^-1.
1. From part (II)(1), we have:
f^-1 = p
Where p: {a,b,c}→{1,2,3}
• In set form, we can write:
f^-1 = {(a,1), (b,2), (c,3)}


2. From part (II)(2), we have:
g^-1 = q
Where q: {apple, ball, cat}→{a,b,c}
• In set form, we can write:
q^-1 = {(apple,a), (ball,b), (cat,c)}

3. From part (II)(3), we have:
(g∘f)^-1 = r
Where r: {apple, ball, cat}→{1,2,3}
• In set form, we can write:
(g∘f)^-1 = {(apple,1), (ball,2), (cat,3)}

4. Now we can calculate f^-1∘g^-1.
(f^-1∘g^-1)(x) = f^-1(g^-1(x))
• For (f^-1∘g^-1),
   ♦ domain is the domain of g^-1 which is {apple, ball, cat}
   ♦ codomain is the codomain of f^-1 which is {1,2,3}
• So we get:
(f^-1∘g^-1)(apple) = f^-1(g^-1(apple)) = f^-1(a) = 1
(f^-1∘g^-1)(ball) = f^-1(g^-1(ball)) = f^-1(b) = 2
(f^-1∘g^-1)(cat) = f^-1(g^-1(cat)) = f^-1(c) = 3

5. We can write the result in (4), in the form of a set:
(f^-1∘g^-1) = {(apple,1), (ball,2), (cat,3)}

6. Comparing the results in (3) and (5), we get:
(g∘f)^-1 = (f^-1∘g^-1)

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The result obtained in the above solved example 17.27 can be used in general. We can write:
If f: X→Y and g: Y→Z are two invertible functions, then (g∘f) is also invertible. The inverse of (g∘f) is (f^-1∘g^-1)

The proof can be written in 7 steps:
1. First we will write some basic details:
(i) Given that, f: X→Y is an invertible function.
• So we can write:
    ♦ f(x) = y
    ♦ f^-1(y) = x
(ii) Given that, g: Y→Z is an invertible function.
• So we can write:
    ♦ g(y) = z
    ♦ g^-1(z) = y
(iii) Note that, domain of g is same as codomain of f.

2. Now we recall an important property of inverse functions.
• If the inverse of f is f^-1, then two conditions will be satisfied:
(i) (f^-1∘f) = I_X
(ii) (f∘f^-1) = I_Y
Where
    ♦ X is the domain of f
    ♦ Y is the domain of f^-1 

• Similarly, if the inverse of g is g^-1, then two conditions will be satisfied:
(i) (g^-1∘g) = I_Y
(ii) (g∘g^-1) = I_Z
Where
    ♦ Y is the domain of g
    ♦ Z is the domain of g^-1

3. Next step is to determine the domain and codomain of the composite functions.
(i) Domain of (g∘f):
• (g∘f) connects the domain of f with the codomain of g.
    ♦ So the domain of (g∘f) is the domain of f, which is X.
    ♦ Also the codomain of (g∘f) is the codomain of g, which is Y. 
(ii) Domain of (f^-1∘g^-1):
• (f^-1∘g^-1) connects the domain of g^-1 with the codomain of f^-1.
    ♦ So the domain of (f^-1∘g^-1) is the domain of g^-1, which is Z.
    ♦ Also the codomain of (f^-1∘g^-1) is the codomain of f^-1, which is X. 

4. Now we can use the property mentioned in step (2).
• That is:
If the inverse of (g∘f) is (f^-1∘g^-1), then two conditions will be satisfied:
(i) (f^-1∘g^-1)∘(g∘f) = I_X
(ii) (g∘f)∘(f^-1∘g^-1) = I_Z
Where
    ♦ X is the domain of (g∘f)
    ♦ Z is the domain of (f^-1∘g^-1)

5. First we show that (f^-1∘g^-1)∘(g∘f) = I_X.
$\begin{array}{ll}{}    &{(f^{-1} \circ g^{-1})(g \circ f)}    & {~=~}    &{\Bigl((f^{-1} \circ g^{-1}) \circ g \Bigr) \circ f ~\color {magenta}{\text{- - - (A)}}}    &{} \\
{}    &{}    & {~=~}    &{\Bigl(f^{-1} \circ (g^{-1} \circ g) \Bigr) \circ f ~\color {magenta}{\text{- - - (B)}}}    &{} \\
{}    &{}    & {~=~}    &{(f^{-1} \circ I_{Y}) \circ f ~\color {magenta}{\text{- - - (C)}}}    &{} \\
{}    &{}    & {~=~}    &{I_X ~\color {magenta}{\text{- - - (D)}}}    &{} \\
\end{array}               
$

◼ Remarks:
• Line marked as A:
Here we use the formula: h∘(g∘f) = (h∘g)∘f
• Line marked as B:
Here also, we use the formula: h∘(g∘f) = (h∘g)∘f
• Line marked as C:
Here we use the fact that: (g^-1∘g) = I_Y
• Line marked as D:
We must simplify (f^-1∘I_Y)∘f. This can be done in two steps:
(i) First we will determine (f^-1∘I_Y)
• Here, the output of I_Y is used as the input for f^-1.
• Also this function connects the domain of I_Y with the codomain of f^-1.
    ♦ Domain of I_Y is Y
    ♦ Codomain of f^-1 is X
So we get: (f^-1∘I_Y) = f^-1(I_Y) = f^-1(y) = x.
(ii) Now we can determine (f^-1∘I_Y)∘f
• Here, the output of f is used as the input for (f^-1∘I_Y).
• Also this function connects the domain of f with the codomain of (f^-1∘I_Y).
    ♦ Domain of f is X
    ♦ Codomain of (f^-1∘I_Y) is also X
So we get: (f^-1∘I_Y)∘f = I_X.

6. In a similar way, we can show that:
(g∘f)∘(f^-1∘g^-1) = I_Z.

7. So both conditions mentioned in (4) are satisfied.
• Therefore it is proved that:
If f: X→Y and g: Y→Z are two invertible functions, then (g∘f) is also invertible. The inverse of (g∘f) is (f^-1∘g^-1)  

Solved example 17.28
Let S = {1,2,3}. Determine whether the functions f: S→S defined as below have inverses. Find f^-1 if it exists.
(a) f = {(1,1), (2,2), (3,3)}
(b) f = {(1,2), (2,1), (3,1)}
(c) f = {(1,3), (3,2), (2,1)}
Solution:
Part (a): f = {(1,1), (2,2), (3,3)}
• Here f is both one-one and onto. So f^-1 exists.
• We can write: f^-1 = {(1,1), (2,2), (3,3)}

Part (b): f = {(1,2), (2,1), (3,1)}
• Here f(2) = f(3) = 1.
So f is not one-one and onto. Therefore, f^-1 does not exist.

Part (c): f = {(1,3), (3,2), (2,1)}
• Here f is both one-one and onto. So f^-1 exists.
• We can write: f^-1 = {(3,1), (2,3), (1,2)}

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Link to a few more solved examples is given below:

Exercise 17.3

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In the next section, we will see binary operations.

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17.7 - Composite of Composite Functions

In the previous section, we saw invertible functions. We saw some solved examples also. In this section, we will see a few more solved examples. Based on those solved examples, we will see composite of composite functions.

Solved example 17.25

Let f: N→R be a function defined as f(x)=4x²+12x+15. Show that f: N→S, where, S is the range of f, is invertible. Find the inverse of f.  
Solution:
1. Based on the given data, we can write 5 points:
(i) f is a set. It contains ordered pairs of the form (x,y).
(ii) The domain of f is N. So x can be natural numbers only.
(iii) For every x, the corresponding y can be calculated using the equation y = 4x²+12x+15
(iv) Since y is calculated using this equation, it will also be a natural number.
(v) The y values thus calculated, forms the set S. Set S is the codomain. In this problem, it is specifically given that, S is the range. So codomain is the range. Recall that, range is a subset of codomain. Range will contain only the outputs.  

2. Now we have the details about f. So we can write the reverse of f. We will call it g.
g: S→N
• The details about g can be written in 5 steps:
(i) g is a set. It contains ordered pairs of the form (y,x)
(ii) The domain of g is set S. The set S is already formed when function f is defined.
(iii) Each element y in S was derived from an element x in N.
• So when we work in reverse from S to N, we can write:
Every element y in S will have a corresponding element x in N.
(iv) y was derived using the equation: y = 4x²+12x+15
We can bring x to the left side as follows:

$$\begin{array}{ll}{}    &{y}    & {~=~}    &{4x^2 + 12x + 15}    &{} \\
{\implies}    &{y}    & {~=~}    &{4x^2 + 12x + 9 + 6}    &{} \\
{\implies}    &{y}    & {~=~}    &{(2x + 3)^2 + 6}    &{} \\
{\implies}    &{(2x+3)^2}    & {~=~}    &{y-6}    &{} \\
{\implies}    &{2x+3}    & {~=~}    &{\sqrt{y-6}}    &{} \\
{\implies}    &{x}    & {~=~}    &{\frac{\sqrt{y-6}~-~3}{2}~\color{magenta}{\text{- - - (I)}}}    &{} \\
\end{array}               
$$

---

◼ Remarks:
• Line marked as I:
Here we must note two points (a) and (b):
(a) Input x must be natural numbers. So the smallest input possible is ‘1’.
• We have y = 4x²+12x+15.
So the smallest possible output corresponding to the smallest possible input will be:
y = 4(1)²+12(1)+15 = 31.
• This value of y is greater than ‘6’. So (y-6) will never become -ve. Therefore we can apply the square root $\sqrt{y-6}$.

(b) When we apply the square root, we usually write: $\pm \sqrt{y-6}$
• But in the present case, the -ve sign is not applicable because, x is a natural number. It cannot be -ve.

---

• So when working in reverse, the x corresponding to y can be obtained as: $x = \frac{\sqrt{y-6}~-~3}{2}$
(v) Based on this, we can define g as follows:
g: S→N is defined as: $x~=~g(y)~=~ \frac{\sqrt{y-6}~-~3}{2}$
• The inputs for g are taken from S. The outputs will be present in N.

3. Now we have both f and g. We can calculate (g∘f)
• (g∘f) means, output of f is used as the input of g.
• The two functions are: f: N→S and g: S→N
• We know that, (g∘f) directly connects the domain of f to the codomain of g. So (g∘f) is from N to N.
• We denoted the inputs from N as x. So we are calculating (g∘f)(x).
• We get:
$$\begin{array}{ll}{}    &{(g \circ f)(x)}    & {~=~}    &{g(f(x))}    &{} \\
{}    &{}    & {~=~}    &{g(4x^2 + 12x + 15)}    &{} \\
{}    &{}    & {~=~}    &{\frac{\sqrt{(4x^2 + 12x + 15) -6}~-~3}{2}}    &{} \\
{}    &{}    & {~=~}    &{\frac{\sqrt{(4x^2 + 12x + 9)}~-~3}{2}}    &{} \\
{}    &{}    & {~=~}    &{\frac{\sqrt{(2x + 3)^2}~-~3}{2}}    &{} \\
{}    &{}    & {~=~}    &{\frac{(2x + 3)~-~3}{2}}    &{} \\
{}    &{}    & {~=~}    &{\frac{2x}{2}}    &{} \\
{}    &{}    & {~=~}    &{x}    &{} \\
\end{array}               
$$

• So whatever input x we give, the output will be the same x. That means, (g∘f) is an identity function.
• The inputs are taken from N. So we can write:
(g∘f) = I_N.

4. Similarly, we can calculate (f∘g)
• (f∘g) means, output of g is used as the input of f.
• The two functions are: g: S→N and f: N→S
• We know that, (f∘g) directly connects the domain of g to the codomain of f. So (f∘g) is from S to S.
• We denoted the inputs from S as y. So we are calculating (f∘g)(y).
• We get:
$$\begin{array}{ll}{}    &{(f \circ g)(y)}    & {~=~}    &{f(g(y))}    &{} \\
{}    &{}    & {~=~}    &{f \left(\frac{\sqrt{y-6} - 3}{2} \right)}    &{} \\
{}    &{}    & {~=~}    &{4 \left[\frac{\sqrt{y-6} - 3}{2} \right]^2 ~+~12 \left[\frac{\sqrt{y-6} - 3}{2} \right] ~+~15 }    &{} \\
{}    &{}    & {~=~}    &{(\sqrt{y-6} - 3)^2 ~+~6(\sqrt{y-6} - 3) ~+~15 }    &{} \\
{}    &{}    & {~=~}    &{(\sqrt{y-6})^2 ~-~ 6 \sqrt{y-6} ~+~ 9 ~+~ 6\sqrt{y-6} ~-~ 18 ~+~15 }    &{} \\
{}    &{}    & {~=~}    &{y-6 ~-~ 6 \sqrt{y-6} ~+~ 9 ~+~ 6\sqrt{y-6} ~-~ 18 ~+~15 }    &{} \\
{}    &{}    & {~=~}    &{y-6 ~+~ 9  ~-~ 18 ~+~15 }    &{} \\
{}    &{}    & {~=~}    &{y}    &{} \\
\end{array}               
$$
• So whatever input y we give, the output will be the same y. That means, (f∘g) is an identity function.
• The inputs for (f∘g) are taken from Y. So we can write:
(f∘g) = I_Y.

5. Let us write a summary:
• We are given a function f: N → S
• We wrote the reverse function g: S → N
• From (3), we got: (g∘f) = I_X
• From (4), we got: (f∘g) = I_Y
• So f is invertible.
• Also, the inverse of f = f^-1 = g
• That means:
The inverse of $y = f(x) = 4x^2 + 12x + 15$ is:
$x = g(y) = \frac{\sqrt{y-6}~-~3}{2}$

Solved example 17.26
Consider f: N→N, g: N→N and h: N→R defined as:
f(x) = 2x,
g(y) = 3y+4 and
h(z) = sin z
for all x, y and z in N.
Show that h∘(g∘f) = (h∘g)∘f.
Solution:
1. (g∘f) = g(f(x)) = g(2x) = 3(2x) + 4 = 6x+4
• (g∘f) connects the domain of f with the codomain of g. So (g∘f) is from N to N.
• The inputs taken from the domain of f are denoted as x.
• So we can write:
(g∘f): N→N is defined as:
(g∘f)(x) = 6x + 4 for all x∈N

2. Now we can find h∘(g∘f)
h∘(g∘f) = h((g∘f)(x)) = h(6x+4) = sin(6x+4)
• h∘(g∘f) connects the domain of (g∘f) with the codomain of h. So h∘(g∘f) is from N to R.
• The inputs taken from the domain of (g∘f) are denoted as x.
• So we can write:
h∘(g∘f): N→R is defined as:
(h∘(g∘f))(x) = sin(6x + 4) for all x∈N

3. (h∘g) = h(g(y)) = h(3y+4) = sin(3y+4)
• (h∘g) connects the domain of g with the codomain of h. So (h∘g) is from N to R.
• The inputs taken from the domain of g are denoted as y.
• So we can write:
(h∘g): N→R is defined as:
(h∘g)(y) = sin(3y+4) for all y∈N

4. Now we can find (h∘g)∘f
(h∘g)∘f = (h∘g)(f(x)) = (h∘g)(2x) = sin(3(2x) + 4) = sin(6x+4)
• (h∘g)∘f connects the domain of f with the codomain of (h∘g). So (h∘g)∘f is from N to R.
• The inputs taken from the domain of f are denoted as x.
• So we can write:
(h∘g)∘f: N→R is defined as: (h∘g)∘f(x) = sin(6x + 4) for all x∈N

5. From (2) and (4), we see that:
(h∘(g∘f))(x) = ((h∘g)∘f)(x) for all x∈N.

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The result obtained in the above solved example 17.26 can be used in general. We can write:
If f: X→Y, g: Y→Z and h: Z→S are functions, then
h∘(g∘f) = (h∘g)∘f

The proof can be written in 5 steps:
1. We are given:
    ♦ f: X→Y
    ♦ g: Y→Z
    ♦ h: Z→S
So we can write:
    ♦ f(x) = y
    ♦ g(y) = z
    ♦ h(z) = s

2. We have to prove that: h∘(g∘f) = (h∘g)∘f

3. First we consider the LHS. We will determine (g∘f) in the LHS.
• (g∘f) connects the domain of f with the codomain of g.
• That means, (g∘f) is from X to Z.
• The inputs will be from X. The outputs should be in Z.
• So we get:
(g∘f)(x) = g(f(x)) = g(y) = z

4. Now we can determine h∘(g∘f)
• h∘(g∘f) connects the domain of (g∘f) with the codomain of h.
•  That means, h∘(g∘f) is from X to S.
• The inputs will be from X. The outputs should be in S.
• So we get:
h∘(g∘f)(x) = h((g∘f)(x)) = h(z) = s

5. Next we consider the RHS. We will determine (h∘g) in the RHS.
• (h∘g) connects the domain of g with the codomain of h.
• That means, (h∘g) is from Y to S.
• The inputs will be from Y. The outputs should be in S.
• So we get:
(h∘g)(y) = h(g(y)) = h(z) = s

6. Now we can determine (h∘g)∘f
• (h∘g)∘f connects the domain of f with the codomain of (h∘g).
•  That means, (h∘g)∘f is from X to S.
• The inputs will be from X. The outputs should be in S.
• So we get:
((h∘g)∘f)(x) = (h∘g)(f(x)) = (h∘g)(y) = s

7. Let us compare the results:
• From (4) we get:
h∘(g∘f)(x) = h((g∘f)(x)) = h(z) = s
• From (6) we get:
((h∘g)∘f)(x) = (h∘g)(f(x)) = (h∘g)(y) = s
• Results are the same. So we can write:
h∘(g∘f)(x) = ((h∘g)∘f)(x) for all x∈X.

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In the next section, we will see two more solved examples.

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17.5 - Properties of Composite Functions

In the previous section, we saw composition of functions. We saw some solved examples also. In this section, we will see a few more solved examples. Those solved examples will help us to learn the properties of composition of functions.

Solved example 17.18
Show that if f: A → B and g: B → C are one-one, then g∘f: A → C is also one-one
Solution:
1. In an one-one function, if (g∘f)(x₁) is to be equal to (g∘f)(x₂), then x₁ must be equal to x₂. This condition can be used to prove that, a given function is an one-one function.
2. In our present case, suppose that, (g∘f)(x₁) is equal to (g∘f)(x₂). Then we can write:

$\begin{array}{ll}{}    &{(g \circ f)(x_1)}    & {~=~}    &{(g \circ f)(x_2)}    &{} \\
{\implies}    &{g(f(x_1))}    & {~=~}    &{g(f(x_1))}    &{} \\
{\implies}    &{f(x_1)}    & {~=~}    &{f(x_2)~\color{magenta}{\text{- - - (I)}}}    &{} \\
{\implies}    &{x_1}    & {~=~}    &{x_2~\color{magenta}{\text{- - - (II)}}}    &{} \\
\end{array}$

◼ Remarks:
• Line marked as I:
We get this result because, g is one-one.
• Line marked as II:
We get this result because, f is one-one.

Solved example 17.19
Show  that  if  f: A  →  B  and  g: B  →  C  are  onto,  then  g∘f: A  →  C  is also onto
Solution:
1. Given that g is onto. So for every element z of C, there will be a pre-image y in B.
    ♦ We can write: g(y) = z
2. Given that f is onto. So for every element y of B, there will be a pre-image x in A.
    ♦ We can write: f(x) = y
3. So we get: (f∘g)(x) = g(f(x)) = g(y) = z
• Taking the first and last items, we get: (f∘g)(x) = z
• That means, for every element z of C, there is a pre-image in A.
• Therefore, (g∘f) is an onto function

Solved example 17.20
Consider functions f and g such that composite g∘f is defined and is one-one. Are f and g both necessarily one-one.
Solution:
We will check this using an example. It can be written in 6 steps:
1. We have two sets:
• A = {1,2,3,4}
• B = {1,2,3,4,5,6}
2. We have two functions:
• f: A→B is defined by: f(x) = x
• g: B→B is defined by:
    ♦ g(x) = x, for x = 1,2,3,4
    ♦ g(5) = 5
    ♦ g(6) = 5
3. Based on the above steps, we can draw the Venn diagrams as shown in fig.17.9 below:

Fig.17.9

4. From the Venn diagram, we see that:
• (g∘f) is one-one.
• f is one-one.
• g is not one-one.
5. So we can write:
Even if the composite function (g∘f) is one-one, it is not necessary for both f and g to be one-one.
6. Let us see a comparison between the two solved examples:
(i) In a previous solved example 17.18, we see that:
If both f and g are one-one, then (g∘f) will be one-one.
(ii) In the present solved example 17.20, we see that:
If (g∘f) is one-one, then it is not necessary for both f and g to be one-one.

Solved example 17.21
Are f and g both necessarily onto, if (g∘f) is onto?
Solution:
We will check this using an example. It can be written in 6 steps:
1. We have two sets:
• A = {1,2,3,4}
• B = {1,2,3}
2. We have two functions:
• f: A→A is defined by:
    ♦ f(x) = x, for x = 1,2
    ♦ f(3) = 3
    ♦ f(4) = 3
• g: A→B is defined by:
    ♦ g(x) = x, for x = 1,2
    ♦ g(3) = 3
    ♦ g(4) = 3
3. Based on the above steps, we can draw the Venn diagrams as shown in fig.17.10 below:

Fig.17.10

4. From the Venn diagram, we see that:
• (g∘f) is onto.
• f is not onto.
• g is onto.
5. So we can write:
Even if the composite function (g∘f) is onto, it is not necessary for both f and g to be onto.
6. Let us see a comparison between the two solved examples:
(i) In a previous solved example 17.19, we see that:
If both f and g are onto, then (g∘f) will be onto.
(ii) In the present solved example 17.21, we see that:
If (g∘f) is onto, then it is not necessary for both f and g to be onto.

• In general, it can be proved that:     ♦ If (g∘f) is one-one, then f is one-one.     ♦ If (g∘f) is onto, then g is onto.

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In the next section, we will see Invertible function.

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17.4 - Composition of Functions

In the previous section, we completed a discussion on one-one and onto functions. In this section, we will see Composition of functions.

Composition of functions

Composition of functions can be explained using an example. It can be written in 8 steps:
1. The display board of a textile shop reads: 14% discount on all items.
2. A person wants to buy a shirt. The marked price of the shirt is Rs 400/-. He wants to know whether he can afford it. He has to consider both the discount and the tax. The tax is to be paid at the rate of 8%.
3. For solving this problem, first we consider the discount.
• We know that:
If x is the marked price, then the cost after 14% discount will be 0.86x
• We can write this as a function: f(x) = 0.86x
This function can be used for all items in that shop.
    ♦ Input ‘x’ is the marked price.
    ♦ Output ‘f(x)’ is the cost after discount.
• In our present case, we can put x₁ = 400
So the cost after discount = f(x₁) = 0.86 × 400 = 344
4. Next we consider the tax.
• We know that:
If x is the cost of an item, the payment to be made considering 8% tax will be 1.08x
• We can write this as a new function: g(x) = 1.08x
This function can be used for all items in that shop.
    ♦ Input ‘x’ is the cost after discount.
    ♦ Output ‘g(x)’ is the payment to be made.
    ♦ In our present case, we can put x₂ = 344
So the payment = g(x₂) = 1.08 × 344 = 371.52
5. Note that, we did two separate calculations:
(i) In the first calculation, we found out the cost after discount.
(ii) In the second calculation, we found out the payment to be made after considering the tax.
6. The two calculations can be done together. It can be explained in 4 steps:
(i) Output of the first function is the input of the second function.
• Output of the first function is 0.86x. Here x is the marked price.
(ii) So input of the second function is 0.86x
• The second function is: g(x) = 1.08x
So we get: g(0.86x) = 1.08(0.86x) = 0.9288x
• Check:
0.9288 × 400 = 371.52
This is the same result that we obtained in (4)
(iii) Now we can write the general form:
• From the above steps, it is clear that:
    ♦ Output of f
    ♦ is the
    ♦ Input of g  
• Using symbols, this can be written in two ways:
    ♦ (g∘f)(x)
    ♦ g(f(x))
We can use any one of the above two ways. Both are read as "g of f of x"
(iv) Note that, the small circle between g and f should not be a filled circle. Then it would look like a dot. If it is a dot, then it means multiplication.
7. (g∘f)(x) is actually a function. It can be plotted just like any other function. The plot in our present case is the magenta line in fig.17.6 below:

Fig.17.6

An explanation of the graph can be written in 5 steps:
(i) Draw the three graphs:
    ♦ f(x) = 0.86x
    ♦ g(x) = 1.08x
    ♦ (g∘f)(x) = 0.9288x
(ii) We want the payment amount when the marked price is 400. So draw a vertical cyan dashed line through 400 on the x-axis.
• This vertical line meets f(x) at a point. Draw a horizontal cyan dashed line through this meeting point.
• This horizontal cyan dashed line meets the y-axis at 344. This is the price after discount.
(iii) We want the payment amount when the cost is 344. So draw a vertical brown dashed line through 344 on the x-axis.
• This vertical line meets g(x) at a point. Draw a horizontal brown dashed line through this point.
• This horizontal brown dashed line meets the y-axis at 371.52. This is the payment amount.
(iv) The horizontal cyan dashed line and vertical brown dashed line, together with the axes, form a square. Why?
(v) In practice, we need not plot f(x) and g(x). All we need is the plot of (g∘f)(x).
• This is because:
The vertical dashed line through 400 meets the magenta line at the white dot. The horizontal dashed line through this white dot meets the y-axis at the same 371.52.
8. (g∘f)(x) can be explained using set theory also. It can be written in 5 steps:
(i) We write set M which contains the marked prices of the various items in the shop. This is shown in fig.17.7(a) below:

Fig.17.7

 
(ii) Each element of M becomes input of the function f. The outputs are the costs after discount. These outputs become elements of set D. Let us see an example:
• 735 is an input for function f. We get: f(735) = 0.86 × 735 = 632.1
• So there is an arrow connecting 735 (in set M) and 632.1 (in set D).
• We can define the function f using sets M and D:
f: M→D defined as f(x) = 0.86x
(iii) Each element of D becomes input of the function g. The outputs are the payment amounts. These outputs become elements of set P. Let us see an example:
• 632.1 is an input for function g. We get: g(632.1) = 1.08 × 632.1 = 682.668
• So there is an arrow connecting 632.1 (in set D) and 682.668 (in set P).
• We can define the function g using sets D and P:
g: D→P defined as g(x) = 1.08x
(iv) Instead of applying two separate functions, we can apply the single function (g∘f). This is shown in fig.17.7(b) above. Let us see an example:
• (g∘f)(735) = 0.9288 × 735 = 682.668
We get the same result as above.
• We see an arrow directly connecting 725 (in set M) and 682.668 (in set P).
• We can define the function (g∘f) using sets M and P:
(g∘f): M→P defined as (g∘f)(x) = 0.9288x
• Note that, M is the domain of the first function and P is the codomain of the second function. So we can write:
The composite function connects the domain of the first function to the codomain of the second function.
(v) So it is clear that, by using composite functions, we can obtain the answers in a single step.    

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Solved example 17.15
Let f : {2, 3, 4, 5} → {3, 4, 5, 9} and g : {3, 4, 5, 9} → {7, 11, 15} be functions defined as:
f(2) = 3,
f(3) = 4,
f(4) = f(5) = 5
and
g(3) = g(4) = 7,
g(5) = g(9) = 11.
Find g∘f.
Solution:
1. We can name the given sets as shown below:
A = {2, 3, 4, 5}
B = {3, 4, 5, 9}
C = {7, 11, 15}
• So the functions can be indicated as:
f : A → B
g : B → C
2. The Venn diagram is shown in fig.17.8(a) below:

Fig.17.8

Let us see an example from the diagram.
• Given that, f(3) = 4,
So there is an arrow connecting 3 (in set A) and 4 (in set B).
• Given that, g(4) = 7,
So there is an arrow connecting 4 (in set B) and 7 (in set C).
3. By following the arrows in fig.a, we get the connection between elements in A and C. This is shown in fig.b above.
• So we can write:
(g∘f)(2) = 7
(g∘f)(3) = 7
(g∘f)(4) = 11
(g∘f)(5) = 11

Alternate method:
(g∘f) means, "output of f" is used as "input of g".
1. Given that, f(2) = 3
So output of f for the input 2 is 3
• This 3 is the input for g.
• Given that, g(3) = 7
So when the input for (g∘f) is 2, the output will be 7
We can write: (g∘f)(2) = 7
2. Given that, f(3) = 4
• Given that, g(4) = 7
So (g∘f)(3) = 7
3. Given that, f(4) = 5
• Given that, g(5) = 11
So (g∘f)(4) = 11
4. Given that, f(5) = 5
• Given that, g(5) = 11
So (g∘f)(5) = 11

Solved example 17.16
Find g∘f and f∘g, if f : R → R and g : R → R are given by f(x) = cos x and g(x) = 3x² . Show that g∘f ≠ f∘g.
Solution:
Part (i): Finding g∘f
1. g∘f means, "output of f" is used as "input of g".
2. Output of f is cos x.
So we can write:
g∘f = g(cos x) = 3(cos x)² = 3 cos²x

Part (ii): Finding f∘g
1. f∘g means, "output of g" is used as "input of f".
2. Output of g is 3x².
So we can write:
f∘g = f(3x²) = cos (3x²)

Part (iii): proving that, g∘f ≠ f∘g
1. We have: g∘f = 3 cos²x
Put x = 0. We get:
g∘f = 3 cos²0  = 3 × 1² = 3
2. We have f∘g = cos (3x²)
Put x = 0. We get:
f∘g = cos (3 × 0) = cos 0 = 1
3. So we can write: g∘f ≠ f∘g

Solved example 17.17
Show that,
• if $f: R- \left \{\frac{7}{5} \right\}~\to~R- \left \{\frac{3}{5} \right\}$ is defined by $f(x)~=~\frac{3x+4}{5x-7}$
• and $g: R- \left \{\frac{3}{5} \right\}~\to~R- \left \{\frac{7}{5} \right\}$ is defined by $f(x)~=~\frac{7x+4}{5x-3}$,
• then f∘g = I_A and g∘f = I_B,
• where
    ♦ $A = R- \left \{\frac{3}{5} \right\}$
    ♦ $B = R- \left \{\frac{7}{5} \right\}$
    ♦ $I_A (x) = x, \forall x \in A$
    ♦ $I_B (x) = x, \forall x \in B$
Solution:
Details about identity function can be seen here.

Part (i): To prove that (f∘g)(x) = I_A(x).
1. First we find (f∘g)
• (f∘g) means, "output of g" is used as "input of f".
Output of g is $\frac{7x+4}{5x-3}$.
So we get:
$$\begin{array}{ll}{}    &{(f \circ g)(x)}    & {~=~}    &{f \left(\frac{7x+4}{5x-3} \right)}    &{} \\
{}    &{}    & {~=~}    &{\frac{3\left(\frac{7x+4}{5x-3} \right)+4}{5\left(\frac{7x+4}{5x-3} \right)-3}}    &{} \\
{}    &{}    & {~=~}    &{\frac{21x+12 + 20x – 12}{35x + 20 -35x + 21}}    &{} \\
{}    &{}    & {~=~}    &{\frac{41x}{41}}    &{} \\
{}    &{}    & {~=~}    &{x}    &{} \\
\end{array}               
$$
2. We see that, (f∘g)(x) is an identity function. This is because, what ever input x we give, the output will be same as the input.
• But we need to specify the the set from which the input values can be taken. In other words, we need to specify the domain.
• We wrote that: $(f \circ g)(x)~=~f \left(\frac{7x+4}{5x-3} \right)$
• When x is $\frac{3}{5}$, the denominator will become zero. So x can be any real number other that $\frac{3}{5}$.
• That means, the domain is $R- \left \{\frac{3}{5} \right\}$
• But we are given that $A = R- \left \{\frac{3}{5} \right\}$.
So the domain is set A.
• We can write:
(f∘g)(x) is an identity function whose domain is A.
• Using symbols, we write this as:
(f∘g)(x) = I_A(x).

Part (ii): To prove that (g∘f)(x) = I_B(x)
1. First we find (g∘f)
• (g∘f) means, "output of f" is used as "input of g".
Output of f is $\frac{3x+4}{5x-7}$.
So we get:
$$\begin{array}{ll}{}    &{(g \circ f)(x)}    & {~=~}    &{g \left(\frac{3x+4}{5x-7} \right)}    &{} \\
{}    &{}    & {~=~}    &{\frac{7\left(\frac{3x+4}{5x-7} \right)+4}{5\left(\frac{3x+4}{5x-7} \right)-3}}    &{} \\
{}    &{}    & {~=~}    &{\frac{21x+28 + 20x – 28}{15x + 20 -15x + 21}}    &{} \\
{}    &{}    & {~=~}    &{\frac{41x}{41}}    &{} \\
{}    &{}    & {~=~}    &{x}    &{} \\
\end{array}               
$$
2. We see that, (g∘f)(x) is an identity function. This is because, what ever input x we give, the output will be same as the input.
• But we need to specify the the set from which the input values can be taken. In other words, we need to specify the domain.
• We wrote that: $(g \circ f)(x)~=~g \left(\frac{3x+4}{5x-7} \right)$
• When x is $\frac{7}{5}$, the denominator will become zero. So x can be any real number other that $\frac{7}{5}$.
• That means, the domain is $R- \left \{\frac{7}{5} \right\}$
• But we are given that $B = R- \left \{\frac{7}{5} \right\}$.
So the domain is set B.
• We can write:
(g∘f)(x) is an identity function whose domain is B.
• Using symbols, we write this as:
(g∘f)(x) = I_B(x)

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In the next section, we will see a few more solved examples. We will also see some important properties of composite functions.

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