﻿ Composite and Inverse Functions Worksheet | Problems & Solutions

# Composite and Inverse Functions Worksheet

Composite and Inverse Functions Worksheet
• Page 1
1.
Given $f$($x$) = 7 + 5$x$ - 2$x$2 and $g$($x$) = 2$x$ + 3, evaluate ($f$ o $g$)($x$).
 a. - 4$x$2 + 10$x$ + 11 b. - 8$x$2 - 14$x$ + 4 c. - 4$x$3 + 4$x$2 + 29$x$ + 21 d. - 4$x$2 + 10$x$ + 17 e. 8$x$2 + 34$x$ + 40

#### Solution:

(f o g)(x) = f[g(x)]
[Composition of f(x) and g(x).]

= f [2x + 3]
[Replace g(x) = 2x + 3.]

= 7 + 5(2x + 3) - 2(2x + 3)2
[Replace x with (2x + 3) in f(x).]

= 7 + 10x + 15 - 2(4x2 + 12x + 9)

= 7 + 10x + 15 - 8x2 - 24x - 18

= - 8x2 - 14x + 4
[Simplify.]

2.
Given $f$($x$) = 5$x$2 - 6$x$ and $g$($x$) = 8$x$ - 1, evaluate ($f$ o $g$)($x$).
 a. 320$x$2 - 128$x$ + 11 b. 320$x$2 - 80$x$ - 1 c. 320$x$2 - 32$x$ - 1 d. 40$x$2 - 48$x$ - 1 e. 40$x$3 - 53$x$2 + 6$x$

#### Solution:

(f o g)(x) = f[g(x)]
[Composition of f(x) and g(x).]

= f [8x - 1]
[Replace g(x) = 8x - 1.]

= 5(8x - 1)2 - 6(8x - 1)
[Replace x with (8x - 1) in f(x).]

= 5(64x2 - 16x + 1) - 48x + 6

= 320x2 - 80x + 5 - 48x + 6

= 320x2 - 128x + 11
[Simplify.]

3.
Given $f$ ($x$) = $\sqrt{x+1}$, evaluate ($f$ o $f$)(8).
 a. 6 b. $\sqrt{\sqrt{7}-1}$ c. 3 d. 9 e. 2

#### Solution:

( f o f )(x) = f [f (x)]
[Composition of f (x) and f (x).]

( f o f )(8) = f [f (8)]
[Replace x = 8.]

= f (8+1)

= f (9) = f (3)

= 3+1 = 4 = 2
[Replace x = 3 in f (x).]

4.
Given $h$($x$) = $x$2 - 9, evaluate ($h$ o $h$)($\sqrt{13}$).
 a. 493 b. 16 c. 160 d. 7 e. 169

#### Solution:

(h o h)(x) = h[h(x)]
[Composition of h(x) and h(x).]

(h o h)(13) = h[h(13)]

= h[(13)2 - 9]
[Replace x with 13 in h(x).]

= h[13 - 9] = h(4)
[Simplify.]

= (4)2 - 9
[Replace x with 4 in h(x).]

= 7
[Simplify.]

5.
Given $g$($x$) = 4$x$ - 3 and $h$($x$) = $\frac{x}{5}$ + $\frac{2}{5}$, evaluate ($g$ o $h$)(- 3).
 a. - $\frac{74}{5}$ b. - $\frac{19}{5}$ c. - $\frac{13}{5}$ d. 1 e. 3

#### Solution:

(g o h)(x) = g[h(x)]
[Composition of g(x) and h(x).]

= g [x5 + 2 / 5]
[Replace h(x) = x5 + 2 / 5.]

(g o h)(- 3) = g[- 3 / 5 + 2 / 5]
[Replace x with - 3.]

= g[- 1 / 5]
[Simplify.]

= 4(- 1 / 5) - 3 = - 4 / 5 - 3
[Replace x = - 1 / 5 in g(x).]

= - 195
[Simplify.]

6.
Given $f$($x$) = 2$x$4 + 5 and $g$($x$) = $x$2 - 7, evaluate ($g$ o $f$)(0).
 a. - $\frac{7}{5}$ b. 18 c. 4807 d. - 12 e. - 35

#### Solution:

(g o f )(x) = g[f(x)]
[Composition of g(x) and f(x).]

= g[2x4 + 5]
[Replace f(x) = 2x4 + 5.]

(g o f )(0) = g[2(0)4 + 5]
[Replace x with 0.]

= g(5)
[Simplify.]

= (5)2 - 7 = 25 - 7 = 18
[Replace x = 5 in g(x).]

7.
Given $g$($x$) = $x$3 and $h$($x$) = 5$x$, evaluate ($h$ o $g$)(- 2).
 a. 2 b. 80 c. - 2 d. - 40 e. $\frac{5}{4}$

#### Solution:

(h o g)(x) = h[g(x)]
[Composition of the functions h(x) and g(x).]

(h o g)(- 2) = h[g(- 2)]
[Replace x = - 2.]

= h[(- 2)3]
[Replace x = - 2 in g(x) = x3.]

= h(- 8)
[Simplify.]

= 5(- 8) = - 40
[Replace x = - 8 in h(x) = 5x.]

8.
Given $f$ ($x$) = 7$x$ - $k$ and $h$($x$) = $\frac{x+9}{4}$, find out for what value of $k$ is ( $f$ o $h$ )($x$) = ( $h$ o $f$ )($x$)?
 a. - 24 b. - 18 c. 24 d. 54 e. 18

#### Solution:

( f o h )(x) = f [h(x)]
[Composition of functions f (x) and h(x).]

= f [x+94]
[Replace h(x) = x+94.]

= 7(x+94) - k = 7x+634 - k
[Replace x = x+94 in f(x).]

( h o f )(x) = h[ f (x)]
[Composition of h(x) and f(x).]

= h[7x - k]
[Replace f(x) = 7x - k.]

= 7x-k+94
[Replace x = 7x - k in h(x).]

( f o h )(x) = ( h o f )(x)
[Equate ( f o h )(x) and ( h o f )(x) to find k.]

7x+634 - k = 7x-k+94

7x+63-4k4 = 7x-k+94
[Simplify.]

- 3k + 54 = 0

3k = 54 k = 18
[Solve for k.]

9.
Given $h$($x$) = $\frac{x}{x+1}$, evaluate ($h$ o $h$ o $h$)(5).
 a. $\frac{5}{11}$ b. $\frac{6}{11}$ c. $\frac{5}{6}$ d. $\frac{5}{16}$ e. $\frac{11}{16}$

#### Solution:

(h o h o h)(x) = (h o h) [h(x)]

= (h o h)(xx+1)
[Composition of (h o h)(x) and h(x).]

(h o h o h)(5) = (h o h)(5 / 6)
[Replace x with 5 and simplify.]

= h[h(5 / 6)]
[Composition of h(x) and h(x).]

= h[5656+1]
[Replace x = 5 / 6 in h(x) = xx+1.]

= h(5 / 11)
[Simplify.]

= 511511+1 = 5 / 16
[Replace x = 5 / 11 in h(x) = xx+1.]

10.
Which of the following is true for $f$ ($x$) = 9$x$ and $h$($x$) = $\frac{2-x}{{x}^{2}}$?
 a. (h o f)(x) = h(x) b. (f o f)(x) = f(x) c. (f o h)(x) ≠ (h o f)(x) d. (f o h)(x) = (h o f)(x) e. (f o h)(x) = f(x)

#### Solution:

( f o h )(x) = f [h(x)]
[Composition of f(x) and h(x).]

= f [2-xx2] = 9[2-xx2]
[Replace h(x) = 2-xx2 in f(x).]

= 18-9xx2f(x)
[Simplify.]

( h o f )(x) = h[f(x)]
[Composition of h(x) and f(x).]

= h[9x] = 2-9x81x2h(x)
[Replace f (x) = 9x and replace x = 9x in h(x).]

(f o f)(x) = f [f(x)]
[Composition of f(x) and f(x).]

= f[9x] = 9(9x) = 81xf(x)

So, ( f o h )(x) ≠ ( h o f )(x).