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If $ f $ and $ g $ are the functions whose graphs are shown, let $ u(x) = f(x) g(x) $ and $ v(x) = f(x)/g(x). $

(a) Find $ u'(1). $

(b) Find $ v'(5). $

(a) 0

(b) $-\frac{20}{27}$

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Oregon State University

Harvey Mudd College

University of Michigan - Ann Arbor

Boston College

Hey, it's clear. So in me rain here. So according to a graph, we could see that F of X is equal to negative one. Have X if X is less than or equal to zero two of X if X is between zero and two included and negative to fifth X minus two plus four. If X is bigger than two than me, ask the derivative, which is equal to negative 1/2 if X is less than zero too, if access between X zero and two and negative to fifth if X is bigger than two and we have G of X, which is equal to to minus X. If X is less than or equal to two than 3/5 times X minus two. If X is bigger than two, we have the derivative of G of X this equal to one negative one. If X is less than two and 3/5 if X is bigger than two. So for party we're gonna find for the derivative of you of one which is equal to the derivative of F of one times the derivative of G of one bless the derivative of up of one times the derivative of G of one. We plug this in to get two times one plus two times negative one. This becomes equal to zero for part B. We're finding for the derivative of V of five. Here we have G of five times the derivative of five minus F five times too derivative of G of five, all over G of five square. You put this in you get 9/5 times and negative to fifth, minus 14 over five times 3 15 a little over 9/5 square, which is equal to negative 20 over 27.