FUNCTION NOTATION FOR SAT

Problem 1 :

If f(x) = 2x√x, then for which of the following values of x does f(x) = x?

A) 1/4     B) 1/2     C) 2     D) 4      E) 8

Solution :

f(x) = 2x√x

Substitute f(x) = x into f(x) = 2x√x

2x√x = x

Take square on both sides

(2x√x)² = (x)²

4x² × x = x²

4x³ = x²

4x³ - x² = 0

x² (4x - 1) = 0

x = 0 or 4x - 1 = 0

x = 0 or x = 1/4

So, option (A) is correct.

Problem 2 :

If f(a) = a^-3 - a^-2, then f(1/3) =

A) -1/6          B) 1/6      C) 6       D) 9       E) 18

Solution :

f(a) = a^-3 - a^-2

Substitute 1/3 for a, then simplify and combine terms.

f(1/3) = (1/3)^-3 - (1/3)^-2

= 1/(1/3)³ - 1/(1/3)²

= 1/(1/27) - 1/(1/9)

= 27 - 9

f(1/3) = 18

So, option (E) is correct.

Problem 3 :

If f(x) = x² + 3x - 4, then f(2 + a) =

A) a² + 7a + 6     B) 2a² - 7a - 12      C) a² + 12a + 3

D) 6a² + 3a + 7       E) a² - a + 6

Solution :

f(x) = x² + 3x - 4

Substitute x = 2 + a into f(x) = x² + 3x - 4

f(2 + a) = (2 + a)² + 3(2 + a) - 4

By using distributive property,

= 4 + 4a + a² + 6 + 3a - 4

f(2 + a) = a² + 7a + 6

So, option (A) is correct.

Problem 4 :

If f(x) = x² and g(x) = x + 3, then g(f(x)) =

A) x + 3     B) x² + 6     C) x + 9     D) x² + 3      E) x³ + 3x²

Solution :

f(x) = x²

g(x) = x + 3

g(f(x)) = f(x) + 3

Substitute x² for f(x)

g(f(x)) = x² + 3

So, option (D) is correct.

Problem 5 :

If f(x) = x/2, then f(x²) √ (f(x))² =

A) x³/4       B) 1          C) 2x²       D) 2        E) 2x  

Solution :

f(x) = x/2

f(x²) √ (f(x))² = (x²/2) √ (x²/4)

= (x²/2) (x/2)

f(x²) √ (f(x))² = x³/4

So, option (A) is correct.

Problem 6 :

If f(x) = √x + 1, and if the domain of x is the set {3, 8, 15}, then which of the following sets indicates the range of f(x)?

A) {-4, -3, -2, 2, 3, 4}     B) {2, 3, 4}         C) {4, 9, 16}

D) {3, 8, 15}            E) {all real numbers}

Solution :

To determine the function’s range, apply the rule √x + 1 to 3, 8 and 15.

√(3 + 1) = √4 = 2

√(8 + 1) = √9 = 3

√(15 + 1) = √16 = 4

= {2, 3, 4}

So, option (B) is correct.

Problem 7 :

If f(a) = 6a - 4, and if the domain of a consists of all real numbers defined by the inequality -6 < a < 4, then the range of f(a) contains all of the following members EXPECT.

A) -24       B) √1/6      C) 0       D) 4      E) 20

Solution :

To determine the functions range, apply the rule (6a - 4) to 6 and to 4. The range consists of all real numbers between the two results.

6(-6) - 4 = -40

6(4) - 4 = 20

Observing options A, B, C and D all lies in between -40 and 20.

The range of the function can be expressed as the set R = {b|-40 < b < 20}.

So, option (E) is correct.

Problem 8 :

If the range of the function f(x) = x² - 2x - 3 is the set R = {0}, then which of the following sets indicates the largest possible domain of x?

A) {-3}      B) {3}     C) {-1}    D) {3, -1}     E) all real numbers

Solution :

f(x) = x² - 2x - 3

x² - 2x - 3 = 0

(x - 3) (x + 1) = 0

x - 3 = 0 or x + 1 =0

x = 3 or x = -1

The largest possible domain of x is the set {3, -1}.

So, option (D) is correct.

Problem 9 :

If f(x) = √x² - 5x + 6, which of the following indicates the set of all values of x at which the function is NOT defined?

A) {x| x < 3}      B) {x| 2 < x < 3}      C) {x| x < -2}

D) {x| -3 < x < 2}        E) {x| x < -3}

Solution :

f(x) = √x² - 5x + 6

Since the square root of a negative number is NOT defined as a real number, we need x² - 5x + 6 to be less than 0.

x² - 5x + 6 < 0

(x - 2) (x - 3) < 0

The function √x² - 5x + 6 is not defined in the interval 2 < x < 3.

So, option (B) is correct.

Problem 10 :

If f(x) = ∛1/x, then the largest possible domain of x is the set that includes

A) all non zero integers

B) all non negative real numbers

C) all real numbers except 0      D) all positive real numbers

E) all real numbers

Solution:

f(x) = ∛1/x

The function f(x) will become undefined, when x = 0, the function can be defined for any other real number value of x.

(If x > 0, then applying the function yields a positive number, if x < 0, then applying the function yields a negative number).

So, option (C) is correct.