Solutions and full explanations to the college algebra multiple choice questions are presented.

9^{log9(4)}=

__Solution__

Exponential and log functions are inverse of each other. Hence

a^{loga(x)}= x , for all x real and positive.

and therefore

9^{log9(4)}= 4

3^{log3(-5)}=

__Solution__

Since -5 is not in the domain of function log_{3}(x),

3^{log3(-5)}is undefined

If f(x) = -2x^{2}+ 8x - 4, which of the follwoing is true?

A. The maximum value of f(x) is - 4.

B. The graph of f opens upward.

C. The graph of f has no x-intercept

D. f is not a one to one function.

__Solution__

f(x) is a quadratic function and its graph is a parabola that may be intercepted by horizontal lines at two points and therefore is not a one to one function. The answer is D

If f(x) = 5 - 2^{x}, then f^{ -1}(-3) =

__Solution__

Find f^{ -1}(x) and then Find f^{ -1}(- 3)

y = 5 - 2^{x}, given

x = 5 - 2^{y}, interchange x and y

2^{y}= 5 - x , y = log_{2}(5 - x) , solve for y

f^{ -1}(x) = log_{2}(5 - x) , inverse function

f^{ -1}(- 3) = log_{2}(5 -(- 3)) = log_{2}8

= log_{2}(2^{3}) = 3

If log_{x}(3) = 1/4, then x =

__Solution__

Rewrite the given equation in exponential form

log_{x}(3) = 1/4 if and only if x^{(1/4)}= 3

We now solve, for x, the exponential equation obtained above by raising both sides to the power 4.

(x^{(1/4)})^{ 4}= 3^{ 4}

x = 3^{ 4}= 81

If f(x) = -x^{2}+ 1, then f(x + 1) =

__Solution__

Substitute x by x + 1 in the formula of f(x) to obtain f(x + 1).

f(x + 1) = - (x + 1)^{ 2}+ 1

Expand and simplify.

f(x + 1) = - x^{ 2}- 2x - 1 + 1 = - x^{ 2}- 2x

If f(x) = x - 4, then (f_{ o }f)(3) =

__Solution__

(f_{ o }f)(3) = f(f(3)) = f(3 - 4) = f(-1) = - 5

If ln(3x - 2) = 1, then x =

__Solution__

Rewrite given equation in exponential form.

ln(3x - 2) = 1 if and only if e^{ 1}= 3x - 2

Solve e^{ 1}= 3x - 2 for x.

x = (e + 2) / 3

The number of real solutions of (x^{2}+ 1)^{2}+ 2(x^{2}+ 1) - 3 = 0 is equal to

__Solution__

Let u = x^{2}+ 1 and rewrite the given equation in terms of u as follows

u^{ 2}+ 2u - 3 = 0

Factor and solve the above equation

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

two solutions: u = x^{2}+ 1 = - 3 and u = x^{2}+ 1 = 1

Equation x^{2}+ 1 = - 3 has no real solutions. Solve the equation x^{2}+ 1 = 1 for to get

x = 0.

The given equation has one real solution.

If the graph of y = (x - 2)^{2}- 3 is translated 5 units up and 2 units to the right, then the equation of the graph obtained is given by

__Solution__

If the graph of y = f(x) is translated 5 units up, the equation of the new graph is given by

y = f(x) + 5

If the graph of y = f(x) + 5 is translated 2 units to the right, the equation of the new graph is given by

y = f(x - 2) + 5 = ((x - 2) - 2)^{2}- 3 + 5

= (x - 4)^{2}+ 2

If f(x) = -e^{x}- 2, then the range of f is given by the interval

A. (-∞ , -2)

B. (-∞ , +∞)

C. (-2; , +∞)

D. (-∞ , +2)

If f(x) = √(x - 1) / (x^{2}- 9), then the domain of f is given by the interval

A. (1 , +∞)

B. (-3 , +3)

C. [1 , 3)U(3 , +∞)

D. (-3 , 3)U(3 , +∞)

The number of points of intersections of the graphs of y = 2^{x}and y = -x^{2}+ 2 is equal to

A. 0

B. 1

C. 2

D. 3

If f(x) = ln(x + 1) - 2, then f^{-1}(x) =

A. e^{x + 1}- 2

B. e^{x}- 2

C. e^{x + 2}- 2

D. e^{x + 2}- 1

For all x real, √(x^{2}-4x + 4) =

A. x - 2

B. x + 2x + 2

C. |x - 2|

D. x + 2

The value of x that makes x^{2}+ 6x + 13 maximum is equal to

A. 6

B. -3

C. 13

D. 3

e^{ln(3) - ln(2) + ln(1/x)}=

A. 3 / (2x)

B. 3x/2

C. 1 + 1/x

D. 3/2 - 1/x

If f(x) = (x - 1) / (x + 2), then the range of f is given by the interval

A. (-∞ , -2) ∪ (-2 , +∞)

B. (-∞ , 1) ∪ (1 , +∞)

C. (-2; , +∞)

D. (-∞ , 1)

ln((x - 1)^{2}) = 2 ln(x - 1) for all x in the interval

A. (-∞ , +∞)

B. [0 , +∞)

C. (-∞ , 1) ∪ (1 , +∞)

D. (1 , +∞)

Let f(x) = x^{2}+ 2x + 4. Which of the following statements is NOT true?

A. f(x) has a maximum value

B. The graph of f is not a line<

C. The graph of f has no x-intercepts.

D. The graph of f has a y-intercept.Answers to the Above Questions

B

D

D

C

A

B

C

B

A

D

A

C

C

D

C

B

A

B

D

A

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