# EVALUATING FUNCTIONS PRACTICE FOR SAT

Problem 1 :

If f(x) = 3x - 1 and 2f(b) = 28, what is the value of f(2b) ?

Solution :

f(x) = 3x - 1

Evaluating f(b), we get

f(b) = 3(b) - 1

Multiplying by 2 on both sides, we get

2f(b) = 2[3b - 1]

28 = 2[3b - 1]

28/2 = 3b - 1

14 = 3b - 1

3b = 15

dividing by 3 on both sides.

b = 15/3

b = 5

f(2b) = f(10) = 3(10) + 1

= 31

Problem 2 :

The function f is defined by f(x) = (x - 7)2 + 9. If f(a- 2) = 25, what is one possible value of a ?

Solution :

f(x) = (x - 7)2 + 9

f(a - 2) = (a - 2 - 7)2 + 9

25 = (a - 9)2 + 9

25 - 9 = (a - 9)2

(a - 9)2 = 16

a - 9 = ±4

a - 9 = 4 and a - 9 = -4

a = 4 + 9 and a = -4 + 9

a = 13 and a = 5

Problem 3 :

A function f(x) has two properties :

f(a + b) = f(a) - b

f(2) = 10

What is the value of f(5) ?

a)  5     b)  7    c) 9      d)  11

Solution :

f(a + b) = f(a) - b

f(2) = 10

When a = 2

f(2 + b) = f(2) - b

f(2 + b) = 10 - b

When b = 3, then b + 2 = 5

f(2+3) = 10 - 3

f(2+3) = 7

Problem 4 :

If f(x + 1) = 3x + 2, the function f could be defined by which of the following?

a)  f(x) = 3x - 2      b)  f(x) = 3x - 1    c)  f(x) = 3x + 1

d)  f(x) = 3x + 5

Solution :

f(x + 1) = 3x + 2

Let x + 1 = t

Then x = t - 1

f(t - 1) = 3(t - 1) + 2

f(t - 1) = 3t - 3 + 2

f(t - 1) = 3t - 1

f(x) = 3x - 1

So, option b is correct.

Problem 5 :

The function f and g are defined by f(x) = x2+ 2 and g(x) = 4x- 3. If a > 0, for what value of a does g(f(a)) = 41 ?

Solution :

f(x) = x2+ 2 and g(x) = 4x- 3

f(a) = a2 + 2

g(f(a)) = 4(a2 + 2)  - 3

41 = 4(a2 + 2)  - 3

44 = 4(a2 + 2)

11 = a2 + 2

a2 = 9

a = 3 and -3

Since a > 0, then the value of a is 3.

Problem 6 :

The function f is defied by f(x) = (1/2)x + a, where a is a constant. If f(a) = 3, what is the value of f(8) ?

a)  6      b)  7       c)  8        d)  9

Solution :

f(x) = (1/2)x + a

f(a) = 3

f(a) = (1/2) a + a

3 = 3a / 2

6 = 3a

a = 2

f(8) = (1/2)(8) + 2

= 4 + 2

f(8) = 6

Problem 7 :

 x013 y202129

The table above displays several points on the graph of the function f in the xy-plane. Which of the following could be f(x) ?

a)  f(x) = 20x     b) f(x) = x + 20     c)  f(x) = x - 20

d) f(x) = x2 + 20

Solution :

Let us consider option d,

 f(x) = x2 + 20f(0) = 02 + 20f(0) = 20 f(x) = x2 + 20f(1) = 12 + 20f(1) = 21 f(x) = x2 + 20f(3) = 32 + 20f(3) = 29

So, option d is correct.

Problem 8 :

For which of the following functions is it true that f(-3) = f(3) ?

a)  f(x) = 2/x     b)  f(x) = x2/3     c)  f(x) = 3x2+1   d)  f(x) = x+2

Solution :

f(-3) = f(3)

 option a :f(x) = 2/xf(-3) = -2/3f(3) = 2/3 option b :f(x) = x2/3f(-3) = (-3)2/3f(-3) = 3 ----(1)f(3) = (3)2/3f(3) = 3 ----(2)

(1) = (2)

So, option b is correct.

Problem 9 :

The function f is defined by f(x) = 3x + 2 and the function g is defined by g(x) = f(2x) - 1. What is the value of g(10) ?

Solution :

f(x) = 3x + 2

f(2x) = 3(2x) + 2

f(2x) - 1 = 3(2x) + 2 - 1

f(2x) - 1 => 6x + 1

g(x) = 6x + 1

g(10) = 6(10) + 1

= 60 + 1

g(10) = 61

Problem 10 :

If f(x) = (16 + x2)/2x for all x ≠ 0, what is the value of f(-4) ?

Solution :

f(x) = (16 + x2)/2x

f(-4) = (16 + (-4)2)/2(-4)

= (16 + 16)/(-8)

= -4

Problem 11 :

 x 0 1 2 f(x) -2 3 18

Several values of the function f are given in the table above. If f(x) = ax2 + b, where a and b are constants, what is the value of f(3) ?

a)  23        b)  39       c)  43       d) 56

Solution :

f(x) = ax2 + b ----(1)

Choosing two points from the table,

(0, -2) and (1, 3)

f(3) = a(3)2 + b

 -2 = a(0)2 + bb = -2 3 = a(1)2 + b3 = a + ba = 3 - ba = 3 - (-2)a = 5

Applying the value of a and b in (1), we get

f(x) = 5 x2 - 2

f(3) = 5(3)2 - 2

= 45 - 2

f(3) = 43

Problem 12 :

 x-4-20234 f(x)3521648

The table above gives some values for the function f. If g(x) = 2f(x), what is the value of k if g(k) = 8 ?

a)  2         b)  3            c)  4         d) 8

Solution :

g(k) = 8

2f(x) = 8

2f(3) =  2(4) ==> 8

Then the value of k is 3.

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