# Solutions and Explanations to Questions College Algebra - sample 1

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

 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 valueB. 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 Algebra Questions and problemsMore ACT, SAT and Compass practice