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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18.1 - Inverse of cosine Function

In the previous section, we saw sin^-1 function. In this section, we will see cos^-1 function.

Some basics can be written in 8 steps:
1. Consider the cosine function:
f(x) = cos x
• The input x can be any real number. So the domain is R.
• The output lies in the interval [-1,1]. So the codomain is [-1,1]
• We saw the above details in class 11. We saw a neat pictorial representation of the above details in the graph of the cosine function. It is shown again in fig.18.6 below:

Fig.18.6

2. Let us check whether the cosine function is one-one.
• Let input x = $\frac{\pi}{2}$.
    ♦ Then the output will be $f \left(\frac{\pi}{2} \right)~=~\cos \left(\frac{\pi}{2} \right)~=~0 $
• Let input x = $\frac{3 \pi}{2}$.
    ♦ Then the output will be $f \left(\frac{3 \pi}{2} \right)~=~\cos \left(\frac{3 \pi}{2} \right)~=~0 $   
• Let input x = $\frac{-3 \pi}{2}$.
    ♦ Then the output will be $f \left(\frac{-3 \pi}{2} \right)~=~\cos \left(\frac{-3 \pi}{2} \right)~=~0 $

(We can cross check with the graph and confirm that the above inputs and outputs are correct)

• We see that, more than one input values from the domain can give the same output. So the cosine function is not a one-one function.

3. Suppose that, we restrict the input values.
• That is, we take input values only from the set $\left[0, \pi \right]$.
• The codomain is the usual [-1,1]
• Then we will get the green curve shown in fig.18.7 below:

Fig.18.7

• In the green portion, no two inputs will give the same output. So the green portion represents a function which is one-one.

4. Next, we have to prove that, the green portion is onto.
• For that, we can consider any y value from the codomain [-1,1].
• There will be always a x value in $\left[0, \pi \right]$, which will satisfy the equation y = cos x.
• So the green portion is onto.
5. We see that, the green portion is both one-one and onto.
• We can represent this function in the mathematical way:
$\text{cosine}:~ \left[0, \pi \right]~\to~[-1,1]$, defined as f(x) = cos x.
6. If a function is both one-one and onto, the codomain is same as range.
• So we can write:
For this function, the domain is $\left[0, \pi \right]$ and range is [-1,1]
7. We know that, if a function is one-one and onto, it will be invertible. We have seen the properties of inverse functions. Let us apply those properties to our present case. It can be written in 4 steps:
(i) If y = f(x) = cos x is invertible, then there exists a function g such that:
g(y) = x
(ii) The function g will also be one-one and onto.
(iii) The domain of f will be the range of g. So the range of g is $\left[0, \pi \right]$.
(iv) The range of f will be the domain of g. So the domain of g is [-1,1]
(iv) The inverse of sine function is denoted as cos^-1. So we can define the inverse function as:
$\cos^{-1}:~ [-1,1]~\to~\left[0, \pi \right]$, defined as x = g(y) = cos^-1 y.
8. Let us see an example:
• Suppose that, for the inverse function, the input y is $\frac{1}{2}$
• Then we get an equation: $x ~=~\cos^{-1} \left(\frac{1}{2} \right)$
• Our aim is to find x. It can be done in 4 steps:
(i) $x ~=~\cos^{-1} \left(\frac{1}{2} \right)$ is an equation of the form $x ~=~f(y)~=~\cos^{-1} \left(y \right)$
(ii) This is an inverse trigonometric function, where input y = $\frac{1}{2}$.
• Based on the inverse trigonometric function, we can write the original trigonometric function:
$y ~=~ f(x) ~=~\cos x$
• In our present case, it is:
$y~=~\frac{1}{2}~=~\cos x$
(iii) So we have a trigonometric equation:
$\cos x~=~\frac{1}{2}$
• When we solve this equation, we get x.
(iv) We have seen the method for solving trigonometric equations in class 11.
• In the present case, we do not need to write many steps. We already know that, $\cos \left(\frac{\pi}{3} \right)~=~\frac{1}{2}$
• So we can write:
$x ~=~\frac{\pi}{3}$

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The above 8 steps help us to understand the basics about cos^-1 function. Now we will see a few more details. It can be written in 5 steps:
1. We saw that, the cos function is not a one-one function. But to make it one-one, we restricted the domain to $\left[0, \pi \right]$.
2. There are other possible “restricted domains” available.
• $\left[- \pi, 0 \right]$ is shown in magenta color in fig.18.8 below:

Fig.18.8

• $\left[\pi, 2 \pi \right]$ is shown in cyan color in fig,18.8 above.
3. There are infinite number of such restricted domains possible.
• We say that:
    ♦ Each restricted domain gives a corresponding branch of the cos^-1 function.
    ♦ The restricted domain $\left[0, \pi \right]$ gives the principal branch of the cos^-1 function.

4. In class 11, we plotted the cosine function. Now we will plot the inverse. It can be done in 4 steps:
(i) Write the set for the original function f. It must contain a convenient number of ordered pairs.
• $\left(\frac{\pi}{3} , \frac{1}{2}  \right)$ is an example of the ordered pairs in f.
(To get a smooth curve, we must write a large number of ordered pairs)
(ii) Based on set f, we can write set g.
This is done by picking each ordered pair from f and interchanging the positions.
• For example, the point $\left(\frac{\pi}{3} , \frac{1}{2}  \right)$ in the set f will become $\left(\frac{1}{2} , \frac{\pi}{3}  \right)$ in set g.
• Thus we will get the required number of ordered pairs in g.
(iii) Mark each ordered pair of g on the graph paper.
• The first coordinate should be marked along the x-axis.
• The second coordinate should be marked along the y-axis.
(iv) Once all the ordered pairs are marked, draw a smooth curve connecting all the marks.
• The smooth curve is the required graph. It is shown in fig.18.9 below:

Fig.18.9

(v) We see that, the graph is smaller in width but larger in height.
• The graph is smaller in width because:
    ♦ Values to the left of -1 cannot be used as input values.    
    ♦ Also, values to the right of 1 cannot be used as input values.
• The graph is larger in height because:
    ♦ Depending upon the branch, values upto +∞ or –∞ can be obtained as output values.
• The red curve is related to the $\left[0, \pi \right]$ branch.
• The cyan curve is related to the $\left[\pi, 2 \pi \right]$ branch. 
• The magenta curve is related to the $\left[- \pi, 0 \right]$ branch.

5. The cos^-1 function is also known as arc cosine function.

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In the next section, we will see cosec^-1 function.

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Chapter 18 - Inverse Trigonometric Functions

In the previous section, we completed a discussion on relations and functions. In this chapter, we will see inverse trigonometric functions.

Some basics can be written in 8 steps:
1. Consider the sine function:
f(x) = sin x
• The input x can be any real number. So the domain is R.
• The output lies in the interval [-1,1]. So the codomain is [-1,1]
• We saw the above details in class 11. We saw a neat pictorial representation of the above details in the graph of the sine function. It is shown again in fig.18.1 below:

Fig.18.1

2. Let us check whether the sine function is one-one.
• Let input x = $\frac{\pi}{2}$.
    ♦ Then the output will be $f \left(\frac{\pi}{2} \right)~=~\sin \left(\frac{\pi}{2} \right)~=~1 $
• Let input x = $\frac{5 \pi}{2}$.
    ♦ Then the output will be $f \left(\frac{5 \pi}{2} \right)~=~\sin \left(\frac{5 \pi}{2} \right)~=~1 $   
• Let input x = $\frac{-3 \pi}{2}$.
    ♦ Then the output will be $f \left(\frac{-3 \pi}{2} \right)~=~\sin \left(\frac{-3 \pi}{2} \right)~=~1 $

(We can cross check with the graph and confirm that the above inputs and outputs are correct)

• We see that, more than one input values from the domain can give the same output. So the sine function is not a one-one function.

3. Suppose that, we restrict the input values.
• That is, we take input values only from the set $\left[\frac{-\pi}{2}, \frac{\pi}{2} \right]$.
• The codomain is the usual [-1,1]
• Then we will get the green curve shown in fig.18.2 below:

Fig.18.2

• In the green portion, no two inputs will give the same output. So the green portion represents a function which is one-one.

4. Next, we have to prove that, the green portion is onto.
• For that, we can consider any y value from the codomain [-1,1].
• There will be always a x value in $\left[\frac{-\pi}{2}, \frac{\pi}{2} \right]$, which will satisfy the equation y = sin x.
• So the green portion is onto.
5. We see that, the green portion is both one-one and onto.
• We can represent this function in the mathematical way:
$\text{sine}:~ \left[\frac{-\pi}{2}, \frac{\pi}{2} \right]~\to~[-1,1]$, defined as f(x) = sin x.
6. If a function is both one-one and onto, the codomain is same as range.
• So we can write:
For this function, the domain is $\left[\frac{-\pi}{2}, \frac{\pi}{2} \right]$ and range is [-1,1]
7. We know that, if a function is one-one and onto, it will be invertible. We have seen the properties of inverse functions. Let us apply those properties to our present case. It can be written in 4 steps:
(i) If y = f(x) = sin x is invertible, then there exists a function g such that:
g(y) = x
(ii) The function g will also be one-one and onto.
(iii) The domain of f will be the range of g. So the range of g is $\left[\frac{-\pi}{2}, \frac{\pi}{2} \right]$.
(iv) The range of f will be the domain of g. So the domain of g is [-1,1]
(iv) The inverse of sine function is denoted as sin^-1. So we can define the inverse function as:
$\sin^{-1}:~ [-1,1]~\to~\left[\frac{-\pi}{2}, \frac{\pi}{2} \right]$, defined as x = g(y) = sin^-1 y.
8. Let us see an example:
• Suppose that, for the inverse function, the input y is $\frac{1}{2}$
• Then we get an equation: $x ~=~\sin^{-1} \left(\frac{1}{2} \right)$
• Our aim is to find x. It can be done in 4 steps:
(i) $x ~=~\sin^{-1} \left(\frac{1}{2} \right)$ is an equation of the form $x ~=~f(y)~=~\sin^{-1} \left(y \right)$
(ii) This is an inverse trigonometric function, where input y = $\frac{1}{2}$.
• Based on the inverse trigonometric function, we can write the original trigonometric function:
$y ~=~ f(x) ~=~\sin x$
• In our present case, it is:
$y~=~\frac{1}{2}~=~\sin x$
(iii) So we have a trigonometric equation:
$\sin x~=~\frac{1}{2}$
• When we solve this equation, we get x.
(iv) We have seen the method for solving trigonometric equations in class 11.
• In the present case, we do not need to write many steps. We already know that, $\sin \left(\frac{\pi}{6} \right)~=~\frac{1}{2}$
• So we can write:
$x ~=~\frac{\pi}{6}$

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The above 8 steps help us to understand the basics about inverse trigonometric functions. Now we will see a few more details. It can be written in 6 steps:
1. We saw that, the sine function is not a one-one function. But to make it one-one, we restricted the domain to $\left[\frac{- \pi}{2}, \frac{\pi}{2} \right]$.
2. There are other possible “restricted domains” available.
• $\left[\frac{-3 \pi}{2}, \frac{- \pi}{2} \right]$ is shown in magenta color in fig.18.3 below:

Fig.18.3

• $\left[\frac{\pi}{2}, \frac{3 \pi}{2} \right]$ is shown in cyan color in fig,18.3 above.
3. There are infinite number of such restricted domains possible.
• We say that:
    ♦ Each restricted domain gives a corresponding branch of the sin^-1 function.
    ♦ The restricted domain $\left[\frac{- \pi}{2}, \frac{\pi}{2} \right]$ gives the principal branch of the  sin^-1 function.

4. In class 11, we plotted the sine function. Now we will plot the inverse. It can be done in 4 steps:
(i) Write the set for the original function f. It must contain a convenient number of ordered pairs.
• $\left(\frac{\pi}{6} , \frac{1}{2}  \right)$ is an example of the ordered pairs in f.
(To get a smooth curve, we must write a large number of ordered pairs)
(ii) Based on set f, we can write set g.
This is done by picking each ordered pair from f and interchanging the positions.
• For example, the point $\left(\frac{\pi}{6} , \frac{1}{2}  \right)$ in the set f will become $\left(\frac{1}{2} , \frac{\pi}{6}  \right)$ in set g.
• Thus we will get the required number of ordered pairs in g.
(iii) Mark each ordered pair of g on the graph paper.
• The first coordinate should be marked along the x-axis.
• The second coordinate should be marked along the y-axis.
(iv) Once all the ordered pairs are marked, draw a smooth curve connecting all the marks.
• The smooth curve is the required graph. It is shown in fig.18.4 below:

Fi.18.4

(v) We see that, the graph is smaller in width but larger in height.
• The graph is smaller in width because:
    ♦ Values to the left of -1 cannot be used as input values.    
    ♦ Also, values to the right of 1 cannot be used as input values.
• The graph is larger in height because:
    ♦ Depending upon the branch, values upto +∞ or –∞ can be obtained as output values.
• The green curve is related to the $\left[\frac{-\pi}{2}, \frac{\pi}{2} \right]$ principal branch.
• The cyan curve is related to the $\left[\frac{\pi}{2}, \frac{3 \pi}{2} \right]$ branch. 
• The magenta curve is related to the $\left[\frac{-3 \pi}{2}, \frac{\pi}{2} \right]$ branch.

5. Consider any function f for which the inverse g exists.
• The graph of g will be the mirror image of the graph of f.
   ♦ The line with equation y=x will be the mirror line.
• We will prove this in the case of sine function. It can be done in 5 steps:
(i) Consider the grid of the graph in fig.18.4 above.
• The x-axis is divided into equal parts, with each part equal to 1 unit.
• The y-axis is also divided into equal parts, with each part equal to π/2 units.
• π/2 = 3.14/2 = 1.57, which is larger than 1. This is the reason why we see larger divisions along the y-axis.
• For the next graph in fig.18.5 below, we will use π/2 for both the axes.

Fig.18.5

(ii) In fig.18.5 above,
    ♦ The sine function is drawn in red color.
    ♦ The sin^-1 function is drawn in green color.
    ♦ The mirror line f(x) = x is drawn in white color.
(iii) Any two convenient points P and Q are marked on the sine function.
• We need to show that, their mirror images P' and Q' lie on the sin^-1 function.
(iv) First we consider point P.
• From P, drop a perpendicular to the mirror line. Let P₁ be the foot of the perpendicular.
• Extend PP₁ so as to intersect with the green curve. Let P' be the point of intersection.
• We can measure and see that, PP₁ = P₁P'.
• Here we see two facts:
    ♦ PP' is perpendicular to the mirror line.
    ♦ PP₁ = P₁P'.
So P' is the mirror image of P.
• We also see that, the coordinates of P' is the interchanged version of P.
(v) Next we consider point Q.
• From Q, drop a perpendicular to the mirror line. Let Q₁ be the foot of the perpendicular.
• Extend QQ₁ so as to intersect with the green curve. Let Q' be the point of intersection.
• We can measure and see that, QQ₁ = Q₁Q'.
• Here we see two facts:
    ♦ QQ' is perpendicular to the mirror line.
    ♦ QQ₁ = Q₁Q'.
So Q' is the mirror image of Q.
• We also see that, the coordinates of Q' is the interchanged version of Q.
• In this way, we can prove that, any point on sin^-1 function is the mirror image of a corresponding point in the sine function.

6. The sin^-1 function is also known as arcsine function.

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We have completed a basic discussion on the sin^-1 function. In the next section, we will see cos^-1 function.

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17.6 - Invertible Function

In the previous section, we saw properties of composite functions. In this section, we will see invertible functions.

First we will see a solved example.

Solved example 17.22
Let f: {1,2,3}  →  {a,b,c}  be  one-one  and  onto  function given by f(1) = a, f(2) = b and f(3) = c. Show that there exists a function g: {a, b, c} → {1, 2, 3}such that g∘f = I_X and f∘g = I_Y, where, X =  {1, 2, 3} and Y = {a, b, c}.
Solution:
1. We have two sets:
• X = {1,2,3}
• Y = {a,b,c}
2. We have a function which is one-one and onto:
• f: X→Y defined by:
    ♦ f(1) = a
    ♦ f(2) = b
    ♦ f(3) = c
3. We can think about a new function in the reverse order. This function is also one-one and onto:
• g: Y→X defined by:
    ♦ g(a) = 1
    ♦ g(b) = 2
    ♦ g(c) = 3
4. Now we can calculate (g∘f)
• Recall that, (g∘f) means, output of f is used as the input of g.
    ♦ (g∘f)(1) = g(f(1)) = g(a) = 1
    ♦ (g∘f)(2) = g(f(2)) = g(b) = 2
    ♦ (g∘f)(3) = g(f(3)) = g(c) = 3
• We see that:
Whatever input we give for the function (g∘f), the output will be the same. So (g∘f) is an identity function.
• But we must specify the domain for the identity function. We see that, the input values for (g∘f) are taken from the set X. So the domain is X. That means, it is an identity function on set X.
• We can write: (g∘f) = I_X.
5. Similarly, we can calculate (f∘g)
• Recall that, (f∘g) means, output of g is used as the input of f.
    ♦ (f∘g)(a) = f(g(a)) = f(1) = a
    ♦ (f∘g)(b) = f(g(b)) = f(2) = b
    ♦ (f∘g)(c) = f(g(c)) = f(3) = c
• We see that:
Whatever input we give for the function (f∘g), the output will be the same. So (f∘g) is an identity function.
• But we must specify the domain for the identity function. We see that, the input values for (f∘g) are taken from the set Y. So the domain is Y. That means, it is an identity function on set Y.
• We can write: (f∘g) = I_Y.

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The result obtained from the above solved example, can be applied in general. It can be written in 2 steps:
1. We have a function f: X→Y
• The function f, is one-one and onto.
2. Then there exists another function g: Y→X such that:
    ♦ (g∘f) = I_X
    ♦ (f∘g) = I_Y

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The converse is also true. It can be written in 2 steps:
1. There exists two functions: f: X→Y and g: Y→X
• The two functions together satisfy two conditions:
    ♦ (g∘f) = I_X
    ♦ (f∘g) = I_Y
2. Then f is both one-one and onto.

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Now we can write about invertible function. It can be written in 4 steps:
1. We have a function f: X→Y
2. There exists another function g: Y→X such that:
    ♦ (g∘f) = I_X
    ♦ (f∘g) = I_Y
3. Then f is an invertible function.
4. The function g is called inverse of f.
Inverse of f is denoted as: f^-1.
5. Note the subscripts X and Y in step(2)
    ♦ X is the domain of the original function f.
    ♦ Y is the domain of the inverse function g.

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Checking whether a function is invertible or not

This can be written in 4 steps:
1. Consider the condition:
    ♦ (g∘f) = I_X
    ♦ (f∘g) = I_Y
2. This condition can be satisfied only if f is one-one and onto.
3. So we can write:
• If f is invertible, then it will be one-one and onto.
• Conversely, if f is one-one and onto, then it will be invertible.
4. This information can be used to prove a function to be invertible.
• All we need to do is: prove that, it is one-one and onto.
• This is especially helpful in situations where we do not need to find f^-1.

Solved example 17.23
Let f: N → Y be a function defined as f(x) = 4x + 3, where, Y = {y ∈ N : y = 4x + 3 for some x ∈ N}.
Show that f is invertible. Find the inverse.
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 = 4x + 3
(iv) Since y is calculated using this equation, it will also be a natural number.
(v) The y values thus calculated, forms the set Y. Set Y is the codomain.
• The codomain Y will not contain natural numbers like 1, 2, 8, 9 etc., This is because, these numbers do not satisfy the equation y = 4x + 3.
• Since the codomain in this case contain only the outputs, we can say that, codomain is same as the range.

2.Now we have the details about f. So we can write the reverse of f. We will call it g.
g: Y→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 Y. The set Y is already formed when function f is defined.
(iii) Each element y in Y was derived from an element x in N.
• So when we work in reverse from Y to N, we can write:
Every element y in Y will have a corresponding element x in N.
(iv) y was derived using the equation: y = 4x + 3
• So when working in reverse, the x corresponding to y can be obtained as: $x = \frac{y-3}{4}$
(v) Based on this, we can define g as follows:
g: Y→N is defined as: $x~=~g(y)~=~ \frac{y-3}{4}$
• The inputs for g are taken from Y. 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→Y and g: Y→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:
$$\begin{array}{ll}{}    &{(g \circ f)(x)}    & {~=~}    &{g(f(x))}    &{} \\
{}    &{}    & {~=~}    &{g(4x+3)}    &{} \\
{}    &{}    & {~=~}    &{\frac{[(4x+3)-3]}{4}}    &{} \\
{}    &{}    & {~=~}    &{\frac{4x}{4}}    &{} \\
{}    &{}    & {~=~}    &{x}    &{} \\
\end{array}               
$$

• 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: Y→N and f: N→Y
• We know that, (f∘g) directly connects the domain of g to the codomain of f. So (f∘g) is from Y to Y.
• We denoted the inputs from Y as y. So we are calculating (f∘g)(y).
• We get:
$$\begin{array}{ll}{}    &{(f \circ g)(y)}    & {~=~}    &{f(g(y))}    &{} \\
{}    &{}    & {~=~}    &{f \left(\frac{y-3}{4} \right)}    &{} \\
{}    &{}    & {~=~}    &{f \left(4 \times \frac{y-3}{4} + 3 \right)}    &{} \\
{}    &{}    & {~=~}    &{f \left(y-3 + 3\right)}    &{} \\
{}    &{}    & {~=~}    &{y}    &{} \\
\end{array}               
$$
• 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 → Y
• We wrote the reverse function g: Y → 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 $f(x) = 4x + 3$ is:
$g(y) = \frac{y-3}{4}$

Solved example 17.24
Let Y = {n² : n∈N} ⊂ N. Consider f: N→Y as f(n) = n². Show that 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 = x²
(iv) Since y is calculated using this equation, it will also be a natural number.
(v) The y values thus calculated, forms the set Y. Set Y is the codomain.
• The codomain Y will not contain natural numbers like 2, 5, 7, 8 etc., This is because, these numbers do not satisfy the equation y = x², where x is a natural number.
• Since the codomain in this case contain only the outputs, we can say that, codomain is same as the range.

2. Now we have the details about f. So we can write the reverse of f. We will call it g.
g: Y→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 Y. The set Y is already formed when function f is defined.
(iii) Each element y in Y was derived from an element x in N.
• So when we work in reverse from Y to N, we can write:
Every element y in Y will have a corresponding element x in N.
(iv) y was derived using the equation: y = x²
• So when working in reverse, the x corresponding to y can be obtained as: $x = \sqrt{y}$
(v) Based on this, we can define g as follows:
g: Y→N is defined as: $x~=~g(y)~=~ \sqrt{y}$
• The inputs for g are taken from Y. 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→Y and g: Y→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:
$$\begin{array}{ll}{}    &{(g \circ f)(x)}    & {~=~}    &{g(f(x))}    &{} \\
{}    &{}    & {~=~}    &{g(x^2)}    &{} \\
{}    &{}    & {~=~}    &{\sqrt{x^2}}    &{} \\
{}    &{}    & {~=~}    &{x}    &{} \\
\end{array}               
$$

• 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: Y→N and f: N→Y
• We know that, (f∘g) directly connects the domain of g to the codomain of f. So (f∘g) is from Y to Y.
• We denoted the inputs from Y as y. So we are calculating (f∘g)(y).
• We get:
$$\begin{array}{ll}{}    &{(f \circ g)(y)}    & {~=~}    &{f(g(y))}    &{} \\
{}    &{}    & {~=~}    &{f \left(\sqrt{y}\right)}    &{} \\
{}    &{}    & {~=~}    &{y}    &{} \\
\end{array}               
$$
• 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 → Y
• We wrote the reverse function g: Y → 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 $f(x) = x^2$ is:
$g(y) = \sqrt{y}$

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

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