MATH 115 QUIZ4-SAMPLE December 7, 2016

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1 MATH 115 QUIZ4-SAMPLE December 7, 2016 Please review the following problems from your book: Section 4.1: 11 ( true and false) Section 4.1: ( Using table or number line.) Section 4.2: Section The following figure depicts a portion of the graph of a function f. Answer the questions a-m about the function f y = f(x) (a) f ( 3) < 0. T F True. Because the function is decreasing at x = 3 (b) f ( 3) < 0. T F False. Because the function is concave up. (c) f (x) > 0 on interval (0, 2). T F False. The function is decreasing on (0, 2). (d) f (x) < 0 on interval (0, 2). T F True. The function is concave down on (0, 2).

2 (e) f (x) > 0 on interval (4, 7). T F False. The function is decreasing on part of the interval. (f) f (x) < 0 on interval (4, 7). T F True. The function is concave down. (g) f ( 2) > 0. T F True. The function is concave up at that point. (h) f ( 2) > 0 T F False. f ( 2) = 0. (The tangent line is horizontal at x = 2.) (i) f (5) = 0. T F True. The tangent line is horizontal. (j) f (2) does not exists. T F True. A corner. (k) f has an inflection point at x = 1. T F True. The graph has a tangent line at x = 1 and the concavity changes when going through c. (l) f has an inflection point at x = 2. T F False. The concavity changes but the tangent line does not exists at x = 2 because of the corner. Page 2

3 (m) f has an inflection point at x = 8. T F True. Even though f has no derivative at x = 8, it has a tangent line ( a vertical one ) and the concavity changes from one side to the other. 2. Find the relative maxima and relative minima of the function if any. F (x) = 1 3 x3 x 2 3x + 4 F (x) = x 2 2x 3 Critical numbers: x 2 2x 3 = 0 gives x = 1, 3. Method 1:First derivative test. Section 4.1 Find the sign for f (x): f So x = 1 gives a relative max and x = 3 gives a relative min. or Method two: Second derivative test. Section 4.2 Find the second derivative. f (x) = 2x 2 Plug in each of x = 1, 3 to find the sign. f ( 1) = 2 ( 1) 2 = 4 < 0 so concave down ( ). That is, x = 1 gives a relative max. f (3) = 2 (3) 2 = 4 > 0 so concave up ( ). That is, x = 3 is a relative min. Plug in f(x) to find y-value: f( 1) = (1/3)( 1) ( 1) 2 3( 1) + 4 = 17/3 Relative max f(3) = (1/3)(3) 3 (3) 2 3(3) + 4 = 5 relative min 3. Sketch the graph of the function having these properties: f(3) = 4, f (3) = 0, Range of f is [4, 12). f (x) < 0 on (, 0) (6, ) and f (x) > 0 on (0, 6). Also, f(0) = f(6) = 8. First draw any point/s given. Here (3, 4), (0, 8) and (6, 8). critical numbers that are given. x = 3 Concavity: Concave upward on (0, 6) and downward on (, 0) (6, ) Page 3

4 The range tells us how high or low we can go (0, 8) 8. (6, 8) (3, 4) Another set of true/false. (a) If the graph of f is concave upward on (a, b), then the graph of f is concave downward on (a, b). (b) T F True. Graph of f and f are reflection of each other. (c) If the graph of f is concave upward on (a, b), then the graph of f(x) + 2 is concave downward on (a, b). T F False. Graph of f(x) + 2 can be obtained by a vertical shift from graph of f. (d) If graph of f is concave upward on (c, b) and concave downward on (a, c), where a < c < b, then f has an inflection point at c. T F False. Tangent line at c may not exist. (e) A polynomial of degree n (n 3) can have at most (n 2) inflection points. T F True. The degree of f (x) is n 2. Page 4

5 (f) If f (c) = 0, then f has a relative extremum at x = c. T F False. The theorem says that sign of f should change as crossing c to have a extremum. (g) If f is differentiable everywhere, f (c) = 0 and f (c) > 0 then f has a relative maximum at x = c. T F False. It has a relative minimum. Page 5

6 Short Notes: These will not appear on your quiz. These are for you to study. A function is increasing/decreasing on an interval (a, b) if for every two numbers x 1 and x 2 in (a, b), (f(x 1 ) < f(x 2 ))/(f(x 1 ) > f(x 2 )), respectively, whenever x 1 < x 2. Theorem: Detection of increasing/decreasing for a differentiable function on (a, b) 1. If f (x) > 0 for every value of x in an interval (a, b), then f is increasing on (a, b). 2. If f (x) < 0 for every value of x in an interval (a, b), then f is decreasing on (a, b). 3. If f (x) = 0 for every value of x in an interval (a, b), then f is constant on (a, b). Determining the intervals where a function is increasing or decreasing 1. Find all values of x where f (x) = 0 or f is discontinuous or f does not exists. Mark them on a line interval So it will break the number line into intervals. 2. Select a test number c in each interval found and determine the sign of f (c) in that interval. If f (c) > 0 then f is increasing on the interval. If f (c) < 0 then f is decreasing on that interval. A function f has a relative maximum/ minimum, at x = c if there exists an open interval (a, b) containing c such that (f(x) f(c))/ (f(x) f(c)), respectively, for all x in (a, b). A CRITICAL number of a function f is any number x inside an interval in the domain of f such that f (x) = 0 or f (x) does not exists. Detection of relative Minimum/Maximum of a differentiable in enough places function. 1. Find f (x). Set equal to zero, f (x) = 0 or find where f is discontinuous, to find the critical numbers. (x values.) 2. Either use First derivative test: Draw a number line and find the sign of the derivative on different intervals. (a) If f (x) changes sign from + to as we move across a critical number c, then f has a relative maximum at x = c. (b) If f (x) changes sign from to + as we move across a critical number c, then f has a relative minimum at x = c. (c) If f (x) does not changes as we move across a critical number c, then f does not have a relative extremum at x = c. 3. Or use second derivative test: Take the second derivative and evaluate at each critical number c (where f (c) = 0). (a) If f (c) < 0, then f has a relative Maximum at c. (.) (b) If f (c) > 0, then f has a relative Minimum at c. (.) (c) If f (c) = 0 or does not exist, then test is inconclusive. Inflection Point: A point on the graph of a continuous function f where the tangent line exists and where concavity changes. Examples: Page 6

7 c c c Finding inflection points of a twice differentiable in most places function. 1. Compute f (x). 2. Determine the numbers in the domain of f for which f (x) = 0 or f does not exists. 3. Determine the sign of f (x) to the left and right of each number found in previous step. If there is a change in sign of f (x) as we move across x = c, then (c, f(c)) is an inflection point of f given that f has tangent line at c. Page 7