9log9(4) =
Solution
Exponential and log functions are inverse of each other. Hence
aloga(x) = x , for all x real and positive.
and therefore
9log9(4) = 4
3log3(-5) =
Solution
Since -5 is not in the domain of function log3(x),
3log3(-5) is undefined
If f(x) = -2x2 + 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 - 2x, then f -1(-3) =
Solution
Find f -1(x) and then Find f -1(- 3)
y = 5 - 2x , given
x = 5 - 2y , interchange x and y
2y = 5 - x , y = log2(5 - x) , solve for y
f -1(x) = log2(5 - x) , inverse function
f -1(- 3) = log2(5 -(- 3)) = log2 8
= log2(23) = 3
If logx(3) = 1/4, then x =
Solution
Rewrite the given equation in exponential form
logx(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) = -x2 + 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 (x2 + 1)2 + 2(x2 + 1) - 3 = 0 is equal to
Solution
Let u = x2 + 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 = x2 + 1 = - 3 and u = x2 + 1 = 1
Equation x2 + 1 = - 3 has no real solutions. Solve the equation x2 + 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) = -ex - 2, then the range of f is given by the interval
A. (-∞ , -2)
B. (-∞ , +∞)
C. (-2; , +∞)
D. (-∞ , +2)
If f(x) = √(x - 1) / (x2 - 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 = 2x and y = -x2 + 2 is equal to
A. 0
B. 1
C. 2
D. 3
If f(x) = ln(x + 1) - 2, then f-1(x) =
A. ex + 1 - 2
B. ex - 2
C. ex + 2 - 2
D. ex + 2 - 1
For all x real, √(x2 -4x + 4) =
A. x - 2
B. x + 2x + 2
C. |x - 2|
D. x + 2
The value of x that makes x2 + 6x + 13 maximum is equal to
A. 6
B. -3
C. 13
D. 3
eln(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) = x2 + 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.
