Question

Sketch the graph of y=f'(x)

13. Below you are given the graph of the function y(x). Use the graph to answer the questions that follow. y- . O-

Answer 1

y = f(n) 6 (-5,4) + 4 3 (1,2) (3 -L 1 2 3 5 6 22 (3,-3 -3 -2) (-/-324 ast -6+ parts - Now the function ya fin has stree this Thelined line, the horizontal line , and the parabok. first post the inclined line =) the two points the line pan through are equation of line 1 y - 4 y yo - 2-4 4,-yo ^, - -3-65) 9 n yo heme => z -no n-no n-65)

-6 y-s -6 z z y-5 nts =) 2. -3+5 nts yas. -3 (nts » y. -3(nes]+5 hence in first part ② -3(n+sts 2 ya fim - 3 2 dy dn f'n z (-38x(1) to (derivative of (2+5) = 1 -5 to 7, the d'in) hence from In dy da -3 N Now the second part, the horizontal line ③ in this region 1 the egnation of horizontal line 2 hence y ye fm) iz f(n) N 2 an derivative of constant is zew) (3 ton I) (-1 encluded) hence poum M2 dyldna flm flna o

Third pact is a parabola. ( (h,k) the equation of a parabola 2 down on wards facing 2 with vesten (h,k) is (a-h) ²-49 (y-k) here (h, k) (1/2) equations hence our 2 becomes (n-1) 4a (g. 2) how we knowe a point which on two points which the parabola passes are (6-1-3), (3; -3) (3,-3) a) substituting (3-1-441-3-3) -> >> (2)? -4a(-s) 4 = - Ya (-5) a 1/5

hence the equation of the parabola becomes (M-12 - 44 945(4-1) (n-1)2 – 4/5(9-2 (y-2) s(n-1) 27 2 27 2 2 oc 2 2- 5 (n-1) hence Ye fino a Q- 5/4(2-1)² > Now - o- 5/4 ( ( (-)</ dy f'm a on 2 an 5 X 8 (n. - [at- s/n (n-1)] . - Sly dy flom-(-1) For n[ - ] the show. / (+-1)

n. y f(n) 2 7 6 (45) ។ 3 3 1 2 2 3 4 5 6 7 - 7 -6 -4 - 3 (1,0) N - please leave a like if you are happy with the answer -3 2 (-5 to -3) d'(n) = dy an ne-3 to-1) (-1, enchided) d'a) z dy 2 an 2 (-1 to op] f'(o) dy - h (n-1 du which is a straight line at (1) dlm - 5/(-1-13 e ush(-2) . S (-1,5) at d'em -5/2 (1-1) = 0 (1,0)