Calculus Exam Review

University:
Felician University
Course:
MATH-160 | College Algebra
Academic year:

2024
Views:

380

Pages:

7
Author:

GalacticGuru

**Calculus AB** **Name:** Kay **Extrema, Special Theorems, First and Second Derivative Tests** 1. The first derivative of the function f is given by f'(x) = sin(x)/x - 1/2. What are its critical numbers on the open interval (0,5)? note: x=0 is a cp? but not on open (0,5) x = .555, x = 1.849 b/c f'(x) = 0 at these x-values 2. The derivative of f is x³(x+5)(x-1). At which values of x will the graph of f have a relative maximum? A relative minimum? Local max at x=0 b/c f' changes from + to - Local min at x=-5, x=1 b/c f' changes from - to + 3. If f"(x) = (x-5)(x-1)(x+2)³, then the graph of f has inflection points when x=? f" Neg Pos Neg Pos x=-2 f" changes neg to pos x=1 f" changes pos to neg x=5 f" changes neg to pos 4. Let f(x) = 2x²/(x²+4), x ∈ [-5,5]. a. On what interval(s) is f increasing? Justify your conclusion. f'(x) = (x²+4)(4x) - 2x²(2x) / (x²+4)² = 16x / (x²+4)² f'(x) = 0 at x=0 f'(x) > 0 for x > 0 So, f increases on (0,5) b. On what interval(s) is f concave up? Justify your conclusion. f"(x) = (x²+4)²*16 - 16x*2(x²+4) / (x²+4)⁴ = 16(4-3x²) / (x²+4)³ f"(x) = 0 at x = ±2/√3 Pos Neg Pos f"(x) > 0 on (-2/√3, 2/√3) c. At what value(s) of x does f have an inflection point? Justify your conclusion. At x = ±2/√3 f"(x) changes sign at each x-value At x = 2/√3, f"(x) changes from pos to neg Sketch a possible graph that has the following characteristics: f(0) = f(3) = 0 f'(0) = f'(3) = 0 f'(x) > 0 if 0 < x < 3 f'(x) < 0 if x > 3 f''(x) < 0 if x < 3 and x > 4 f''(x) > 0 if 3 < x < 4 Justification for why f has an absolute max @ x=0 f is continuous at x=0 f'(x) = 0 at x=0 f'(x) > 0 for x<0 and f'(x) < 0 for x>0 Let f be the function given by f(x) = -x^(2/3). Which of the following statements about f are true? I. f is continuous at x=0 True II. f is differentiable at x=0 False f'(0) dne III. f has an absolute maximum at x=0 True f'(x) = -2/3x^(-1/3) f''(x) = 2/9x^(-4/3) f'(x) > 0 for x<0 -> f inc f'(x) < 0 for x>0 -> f dec I and III are true The graph of the derivative of the function, f(x), is shown at the right. a. Locate the x-value(s) where there is a relative maximum. Justify. x = 2 f'(x) changes pos to neg b. Locate the x-value(s) where there is a relative minimum. Justify. x = -3, x = 3 f'(x) changes neg to pos a. Find the x-coordinate(s) of the point(s) of inflection of the graph of f. Give a reason for your answer. x=1, x=5 b/c f'(x) has a local extrema f"(x) changes sign x=1 f' has local min x=5 f' has local max b. For what values of x does f attain its absolute minimum value on the closed interval [-2, 8]? Show the analysis that leads to your answer. Candidates: x=-2, 8 (endpoints) Since f'(x) never changes from neg to pos, no critical pts are candidates for abs. min Since f inc (-2, 6) and dec (6,8) the absolute min occurs at x=-2 c. Using the known points given on the graph of f', for what value(s) of x does the graph of y=-1/3x + f(x) have a horizontal tangent? Give a reason for your answer. (-2,4) f'(-2)=4 (0,5) f'(0)=5 (1,0) f'(1)=0 (3,1) f'(3)=1 (5,2) f'(5)=0 (6,0) f'(6)=0 (8,-5) f'(8)=-5 d. Let g be the function defined as g(x) = x²f(x). Find an equation for the line tangent to the graph of g at x=3. g'(x) = x²f'(x) + f(x)*2x g(3) = 9*f(3) = -18 g'(3) = 9*f'(3) + f(3)*6 = 9*1 + (-2)*6 = -3 y + 18 = -3(x - 3)