EVALUATING FUNCTIONS WORD PROBLEMS FOR SAT

Problem 1 :

The volume of a balloon is given by the equation

V = t2 - 3t + 3

What is the volume of the balloon after 3 seconds ?

a)  3    b)  6    c)  9    d)  12

Solution :

V = t2 - 3t + 3

When t = 3

V = 32 - 3(3) + 3

V = 9 - 9 + 3

V = 3

So, volume of balloon after 3 seconds is 3.

Problem 2 :

x

1

2

h

f(x)

0

h

k

In the table above, if f(x) = x2 + x - 2, what is the value of k ?

Solution :

f(x) = x2 + x - 2

When x = 2, f(2) = h

f(2) = 22 + 2 - 2

h = 4

When h = 4, f(4) = k

f(4) = 42 + 4 - 2

k = 16 + 4 - 2

k = 18

Problem 3 :

Rocket

Rocket 1

Rocket 2

Rocket 3

Rocket 4

Rocket 5

Rocket 6

Rocket 7

Fuel burned (liters)

7

12

17

23

29

32

35

The distance d, in meter traveled by a rocket depends on the amount of fuel f, in liters. It burns according to the equation

d = (2/3) f

Based on the table above, how many rockets traveled more than 20 meters ?

a)  One    b)  Two     c)  Three    d)  Four

Solution :

d = (2/3) f

when f = 7

d = (2/3) (7)

d = 14/3 (< 20)

when f = 23

d = (2/3) (23)

d = 15.3 (< 20)

when f = 29

d = (2/3) (29)

d = 19.3 (< 20)

When f = 32

f = (2/3)(32)

f = 21.3 > 20

For f = 32 and f = 35, the value of f will be greater than 20. So, answer is 2.

Problem 4 :

Let the function f be defined by f(x) = 2x3 - 1 and let the function g be defined by g(x) = x2 + 3, what is the value of f(g(1))?

a)  4       b) 23      c)  56    d)  127

Solution :

f(x) = 2x3 - 1

g(x) = x2 + 3

g(1) = 12 + 3 ==> 4

f(g(1)) = f(4) = 2(4)3 - 1

= 2(64) - 1

= 128 - 1

= 127

So, option D is correct.

Problem 5 :

x

1

2

3

4

f(x)

4

6

5

2

g(x)

0

2

6

4

Four values for the function f and g are shown in the table above. If g(m) = 6, what is the value of f(m) ?

Solution :

For g(3), we get 6. So, g(m) = 6. Then value of m is 3.

f(3) = 5

Problem 6 :

f(x) = ax3 + b

In the function f defined above, a and b are constants. If f(-1) = 4 and f(1) = 10, what is the value of b ?

Solution :

f(x) = ax3 + b

f(-1) = a(-1)3 + b 

4 = -a + b ------(1)

f(1) = a(1)3 + b 

10 = a + b ------(2)

(1) + (2)

-a + b + a + b = 4 + 10

2b = 14

b = 7

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

-a + 7 = 4

-a = 4 - 7

-a = -3

a = 3

Problem 7 :

For all x ≥ 3, f(x) = √(x-3)/2 .If f(n) = 3 is the value of n ?

Solution :

f(x) = √(x-3)/2

If f(n) = 3,

f(n) √(n-3)/2

√(n-3)/2

Multiplying by 2 on both sides.

3(2) = √(n - 3)

6 = √(n - 3)

Take square on both sides, we get

36 = n - 3

n = 36 + 3

n = 39

Problem 8 :

Let the function f be defined by f(x) = x2 - x + 3. If f(m + 1) = 5 and m > 0. What is the value of m ?

Solution :

f(x) = x2 - x + 3

f(m + 1) = 5

f(m + 1) = (m+1)2 - (m+1) + 3

5 = m2 + 2m + 1 - m - 1 + 3

5 = m2 + m + 3

m2 + m + 3 - 5 = 0

m2 + m - 2 = 0

(m + 2)(m - 1) = .0

Equating each factor to 0, we get

m  = -2, m = 1

Problem 9 :

f(x) = x3 + 2

g(x) = 2x

The functions f and g are defined above. If f(b) = 29, what is the value of g(b) ?

a)  4    b)  6    c)  8     d) 10

Solution :

f(x) = x3 + 2

f(b) = 29

f(b) = b3 + 2

29 = b3 + 2

b3 = 29 - 2

b3 = 27

b = 3

Applying the value of b in g(x)

g(b) = g(3) = 2(3)

g(3) = 6

Problem 10 :

If f(x) = √(x + 1), what is f(x2 + 4) equal to ?

a) x + 3       b)  x + 5      c) √(x2 + 4) + 1    d) √(x2 + 4x) + 1

Solution :

f(x) = √(x + 1)

f(x2 + 4) √(x2 + 4 + 1)

So, option c is correct.

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