Question 13-4| -3 -2| -11| 23 4 5...S'(x) -1-2 01| 2| 110|-2| -3-1The derivative, f', of a function f is continuous and has exactly two zeros on [-4,5]. Selected valuesof f'(x) are given in the table above. On which of the following intervals is f increasing?(A)-3

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Answered: Question 13 -4|-3 -2| -1 O12 3 4 5 ...…

Big Ideas Math A Bridge To Success Algebra 1: Student Edition 2015

1st Edition

ISBN:9781680331141

Author:HOUGHTON MIFFLIN HARCOURT

Publisher:HOUGHTON MIFFLIN HARCOURT

Chapter8: Graphing Quadratic Functions

Section: Chapter Questions

Problem 17CT

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The function f(x) is continuous and differentiable on the interval [−2,6]. It is known that f(−2)=4 and the derivative on the interval satisfies the condition f′(x)≤3 for all x∈(−2,6). Determine an upper bound of the function at the right endpoint x=6.
QUESTION 2 Using Mean Value Theorem, Prove that if two differentiable functions f, g agree on their first derivatives, that is, f'=g' then the two functions differ only by a constant.
1)Describe the purpose of finding the first and second derivatives to sketch a graph of functions. 2)Find all relative extrema using the second derivative test. a. f(x) = x^2(6 − x)^3b. g(x) = x^3 −5x2 +7x
#12 on the attached picture composition of functions
Let 2x-4/x-1 (d) Determine intervals on which the function is increasing; determine intervals on which the function is decreasing. (h) With the aid of the information obtained in parts (a) - (g), give a reasonable sketch of the curve. (a) X=2 (2,0) Y=4 (0,4) (g) To find the inflection point, we need to find the value for which the double derivative is equal to 0. f''(x)−4(x−1)3−4===000f''(x)=0-4(x-1)3=0-4=0 This statement is false because −4≠0-4≠0 for any real value of x. Since f''(x) cannot be equal to zero for any real value of x, therefore there is no inflection point. Answer: (e) No relative maxima and no relative minima. (f) Function f(x) is concave up over the interval (−∞,1)(-∞,1) and concave down over the interval (1,∞)(1,∞). (g) There is no inflection point. please quickly
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1) Describe the purpose of finding the first and second derivatives to sketch a graph of functions. 2) Find all relative extrema using the second derivative test.a. f(x) = x^2(6 − x)^3b. g(x) = x^3 −5x^2 +7x
h(x) = 4/((x-1)^2) Show from the definition that the function h(x) = 4/((x-1)^2) is increasing in the interval (-∞, 1) and decreasing in the interval (1,+∞).
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Question 13 -4|-3 -2| -1 O12 3 4 5 ... S'(x) -1 -2 2 10|-2| -3-1 The derivative, f' , of a function f is continuous and has exactly two zeros on [-4,5]. Selected values of f'(x) are given in the table above. On which of the following intervals is f increasing? A -3