# Function Worksheets - Page 3

Function Worksheets
• Page 3
21.
If $f$ ($x$) = | $x$ - 13|, then what is the value of $\frac{f\left(14\right)}{f\left(11\right)}$ ?
 a. - $\frac{1}{24}$ b. - $\frac{27}{2}$ c. $\frac{27}{2}$ d. $\frac{1}{2}$

#### Solution:

f (x) = | x - 13 |

f (14) = | 14 - 13 | = 1
[Substitute x = 14 in f(x).]

f (11) = |11 - 13| = |- 2| = 2
[Substitute x = 11 in f(x).]

So, f(14)f(11) = 1 / 2

22.
If $f$ ($x$) = 2$x$ + 4$x$2 + 5$x$3, then find the value of $f$ (2) + $f$ (3) + $f$ (4).
 a. 629 b. 509 c. - 155 d. - 275

#### Solution:

f (x) = 2x + 4x2 + 5x3

f (2) = 2(2) + 4(2)2 + 5(2)3 = 60
[Substitute x = 2 in f (x).]

f (3) = 2(3) + 4(3)2 + 5(3)3 = 177
[Substitute x = 3 in f (x).]

f (4) = 2(4) + 4(4)2 + 5(4)3 = 392
[Substitute x = 4 in f (x).]

So, f (2) + f (3) + f (4) = 629

23.
If $f$($x$) = 6$x$3, then =
 a. 18$x$2 + 18$\mathrm{xh}$ + 6$h$2 b. - 18$x$2 + 18$\mathrm{xh}$ - $h$2 c. - 18$x$2 + 18$\mathrm{xh}$ - 6$h$2 d. 18$x$2 + 18$\mathrm{xh}$ + $h$2

#### Solution:

f(x) = 6x3

f( x + h) = 6(x + h)3
[Substitute (x + h) for x in f(x).]

So, f(x+h)-f(x)h = 6(x + h)3 -  6x3h

= 18x2 + 18xh + 6h2
[Simplify.]

24.
If $f$($x$) is any real function, then which of the following is true?
 a. $f$($x$ + $y$) = $f$($x$) + $f$($y$) b. $f$($x$ - $y$) = $f$($x$) - $f$($y$) c. $f$ 2($x$) = ($f$ ($x$))2 d. None of the above

#### Solution:

The choices A, B and D are not true in the case of any function f(x) always, which can be checked out by taking any random function.

Since f 2(x) is defined as [ f(x)]2, the choice C is true.

25.
If $f$ ($x$) = $x$ + 5, $g$($x$) = ($x$ + 5)2 then evaluate ($f$ + $g$)(3).
 a. 14 b. 72 c. 64 d. 22

#### Solution:

f(x) = x + 5, g(x) = (x + 5)2

f(3) = 3 + 5 = 8
[Substitute x = 3 in f(x).]

g(3) = (3 + 5)2 = 64
[Substitute x = 3 in g(x).]

So, (f + g)(3) = f(3) + g(3) = 8 + 64 = 72
[Use (f + g)(a) = f(a) + g(a).]

26.
If $f$($x$) = $\frac{9}{x}$, $g$($x$) = 2$x$2 then find the value of ($f$ - $g$)(- 3).
 a. 21 b. 15 c. - 21 d. - 15

#### Solution:

f(x) = 9x and g(x) = 2x2

f(- 3) = 9- 3 = - 3
[Substitute x = - 3 in f(x).]

g(- 3) = 2(- 3)2 = 18
[Substitute x = - 3 in g(x).]

So, (f - g)(- 3) = f(- 3) - g(- 3) = - 3 - 18 = - 21
[Use (f - g)(a) = f(a) - g(a). ]

27.
If $g$($x$) = , $h$($x$) = then find the value of ($h$$g$)(8).
 a. $\sqrt{15}$ b. 15 c. 8 d. 1

#### Solution:

g(x) = x + 7and h(x) = x - 7

g(8) = 8 + 7 = 15 = 15
[Substitute x = 8 in g(x).]

h(8) = 8 - 7 = 1 = 1
[Substitute x = 8 in h(x).]

So, (hg)(8) = h(8) g(8) = (1)(15) = 15
[Use (fg)(a) = f(a) × g(a).]

28.
If $f$($x$) = and $g$($x$) = 2 + $f$($x$) then find the value of $\frac{f}{g}$(0).
 a. $\frac{1}{3}$ b. $\frac{1}{4}$ c. $\frac{1}{8}$ d. 2

#### Solution:

f(x) = x - 2x - 3 and g(x) = 2 + f(x)

f(0) = 0 - 20 - 3 = 2 / 3
[Substitute x = 0 in f(x).]

g(0) = 2 + f(0) = 2 + 2 / 3 = 8 / 3
[Substitute x = 0 in g(x).]

So, fg(0) = 2383 = 1 / 4
[Use fg(a) = f(a)g(a).]

29.
If $f$($x$) = , $g$($x$) = and $h$($x$) = then find the value of ()(5).
 a. $\frac{31}{5}$ b. $\frac{51}{6}$ c. $\frac{-63}{2}$ d. $\frac{41}{4}$

#### Solution:

f (x) =6 - x, g(x) = (130 - x)3, h(x) = (630 - x) 4

f (5) = 6 - 5 = 1 = 1
[Substitute x = 5in f(x).]

g (5) = (130 - 5)  3 = 125  3 = 5
[Substitute x = 5 in g(x).]

h (5) = (630 - 5)  4= 625  4 = 5.
[Substitute x = 5 in h(x).]

So, (fg +gh +hf ) (5) = f(5)g(5) +g(5)h(5) + h(5)f(5)
[Use (fg +gh +hf ) (a) = f(a)g(a) +g(a)h(a) + h(a)f(a) .]

= ( 1 / 5) + (5 / 5) + (5 / 1) = 1 / 5 + 1 + 5 = 31 / 5

30.
If f ($x$) = $x$2 + 5$x$ + 9 and $g$($x$) = $x$2 + 7$x$ + 9 then what is the value of $f$$g$(5)?
 a. 59 b. 4071 c. $\frac{59}{69}$ d. 25

#### Solution:

f(x) = x2 + 5x + 9 and g(x) = x2 + 7x + 9

f(5) = (5)2 + 5(5) + 9 = 59
[Substitute x = 5 in f(x).]

g(5) = (5)2 + 7(5) + 9 = 69
[Substitute x = 5 in g(x).]

So, fg(5) = f(5) × g(5) = (59)(69) = 4071
[Use fg(a) = f(a) × g(a).]