Given f'(x)=x²-x 2 and its graph, use the graph to determine the following. f'(x) 10 X = 2' -4 -3 -2 -1 9 ·∞, X 4 2 X relative maxima -41 -6 Find the critical values of f(x). (You may assume that all critical values are integers. Enter your answers as a comma-separated list.) X -10 1 1 2 L 3 Find intervals where f(x) is increasing. (Enter your answer using interval notation.) ( 1217,00) L 4 X Find intervals of where f(x) is decreasing. (Enter your answer using interval notation.) Find where f(x) has relative maxima, relative minima, and horizontal points of inflection. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) x = DNE

Given f'(x)=x²-x 2 and its graph, use the graph to determine the following. f'(x) 10 X = 2' -4 -3 -2 -1 9 ·∞, X 4 2 X relative maxima -41 -6 Find the critical values of f(x). (You may assume that all critical values are integers. Enter your answers as a comma-separated list.) X -10 1 1 2 L 3 Find intervals where f(x) is increasing. (Enter your answer using interval notation.) ( 1217,00) L 4 X Find intervals of where f(x) is decreasing. (Enter your answer using interval notation.) Find where f(x) has relative maxima, relative minima, and horizontal points of inflection. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) x = DNE

Calculus For The Life Sciences

2nd Edition

ISBN:9780321964038

Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Chapter5: Graphs And The Derivative

Section5.1: Increasing And Decreasing Functions

Problem 33E

See similar textbooks

Similar questions

Find all points on the graph of fx=x3+9x2+19x-10 where the slope of the tangent line is -5.
differentiate y = [(x + 11)/(x - 3)]3
Find the zeroes of the function f(x) = x/(x2-1)
- x + 1 If f(x) = find (fof) (x). X - 1
Suppose f(x) = x² + 3x + 1. In this problem, we will show that f has exactly one root (or zero) in the interval [-4,-1]. (a) First, we show that f has a root in the interval (-4,-1). Since f is a choose function on the interval [-4, -1] and f(-4)= and f(-1) =, the graph of y = f(x) must cross the x-axis at some point in the interval (-4,-1) by the ◆. Thus, f has at least choose one root in the interval [—4, -1]. (b) Second, we show that f cannot have more than one root in the interval [−4, -1] by a thought experiment. Suppose that there were two roots x = a and x = b in the interval [-4, −1] with a < b. Then f(a) = f(b) = Since fis choose on the interval [-4,-1] and choose on the interval (-4,-1), by choose there would exist a point c in interval (a, b) so that f'(c) = 0. However, the only solution to f'(x) = 0 is x = not in the interval (a, b), or in [−4, -1]. Thus, f cannot have more than one root in [−4, −1]. 0 which is
Find the interval(s) where the function is increasing and the interval(s) where it is decreasing. (Enter your answers using interval notation. If an answer cannot be expressed as an interval, enter EMPTY or ∅.) f(x) = 1/3x^3 -3/2x^2