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:

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

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◼ 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.

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• 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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