Question

answering D) only will be enough. thank you!

9. Use the graph of f'(x) given below to answer the following questions: á Sobo -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 (a) What is ſ s'(x) dx? J-6 (b) If f(-1) = 4, what is f(-6)? (c) If f(0) = 7, what is f(4)? (d) If f(-2) = 0, what is f (6)?

Answer 1

Og (d) 7 film) D12,2) (-1,2) B1-4,-1) / (49,2 F 15,-3) (-6, -2) to intervals -4+6 clear that f'(al has many definition according () XE (-6,-47: equation of AB is ( f'(x) + 1) = (-1 + 2 ) (x+4) f'(x) + 1 = (x+u) => f'(X) = x/2 + 2-1 = f'(n) = x/2 + 1 еччh cm Вс f'(x) -2 = 12+1 + - +

x-axis? - f(n) = (x+1 1+2 » f'(x) = x+3 XE [1, 2] f'(x) = 2 I line barellel to (iv) a E 22,4) egn of line DE f'(al - 2 = (2+2) (x - 2) - f'(x) = (-2x + 4) + 2 = f'(o) = -x + 6 (v) x + [ 4 , 1 equ of live EF f'lal + 2 = (-2 +3 4-5 f'(x) = |-x+4) - 2 -> f'(x) = -x+2 (vi) XE [5,77 egn of line FG f'(a1 + 3 = -3 -3 (2-5) 5-7 f'in = (3x-15 1-3 f'(a) = 3x - 18

therefore x12 + 1 x + 3 -64 x - y -u< x < -1 , f'(x) = + 3x - 18 ! - 20+ 6 -x+2 , mucas 5 ska < on integration x² + x + c, , -6s as-4 x% +34 + (2, -usas - 2x + (3 , -, sa se - x² + 6x + Cu , 2<x <u -20% + 2a +(5, usx<5 30% – 184 + 6. sexs 7 f(x) = given that f(-2) = 0 (+212 + 3(-2) + C₂ = 0 - 2 - 6 + C2 = 0 C2 - 4 f(x) is differentiable flal is continuous also

il fl-1th) sim +/-1-h) = sim hyo 1/2 + (-3) + 4 = -2+ 3/2 = - 2 + (z C = 2 + 3% = 7/2 v > - I'm fl-h) - lin f (246) hto ho - 4 + 12 + Cy 4+(7/2) = 4+ 12 + cu Cue u+72-8 cu = 1/2 uy su vo lim flurnl = lim flurn) hoo hto = - 8 + 8 - 16 + 24 - + cs cs = 15h also h-o lim f(s-nl = lief (5th) - 25/2 + 10 + 15/2 = 3x25/ - 90 + 6 Co - 10-5 + 90 – 7572 = 95 - 75/2 = 115%. »

therefore f(a) = 3x2 – 18x + 115/2 (16) = 3 x 36 - 1846 + 115/2 , xe [ 57] f (6) = which is 7/2 required value .