Question

(5 points) For the function y = 5x2: (a) Find the average rate of change of y with respect to x over the interval [5,7). (b) Find the instantaneous rate of change of y with respect to x at the value x = 5. Average Rate of Change: | Instantaneous Rate of Change at x = 5:

Answer 1

0 y = 5x² (a) rate of Average n'e (517) change of y wortx Given; 512 f(n) = 5 ca <7 a = 5 b=7 The Average rate of change is, ACR) f(b) -f(a) b-a f(a) f15) - 5(512 -125 f(b) f(7) - 5 (712 - 245 The Average rate change is A (2) = 245 - 125 7-5 120 2 A(n) = 60

(b) instantaneous rate of change of with respect to a at x = 5 q f(0) = f'(x) 5x² 5. d (2) dn f'(o) 5. (20) f'(a) 10 u Thus the instantaneous rate of change at 2 = 5 f'(5) 10 (5) f'(5) 50 f(x) 3x3 + 3n+2 use the imit calculate definition of the derivative f'(n) i.e to f'(n) f(n) lim hao f(ath) h

+00= lim hto [ 3(x+h)3 + 3 (u+h) + 2 +2] - [ 343 +34 +2] h lim hao ( 3 (X3413 + 3 xºh + 3x4°) + 3x+3h+2 B +31 + 2 3 43 +30 +2 h lim ho 3^3 + 343 +9 n'h + 93h2+3x+3h+2-313-31-2 h lim hto 3h3 +9 1²h +gah² + 3h h lim hro h ( 3n²+ 9x² + 9xh+3) h lim hto 3h² + 9x²+ 9 ah + 3 lim hto. 0 + 9x² +0+3 x²+3

Therefore, f(a) 9 U² +3 To calculate second derivative of f(n) i e f"(W) d [fiem] du f'(W) = 9x²+3 let therefore, g(x) g'() = f'(X) = ( 9x+3) dn use of limit definition of to calculate derivative f(x) glin) or g (ath) gren lim hro g() h bìu [ ( 1+ b) + ] - [ x + 3] ho h lim hto g(n++he+24x) + 3] - ( 9**+3) h

lim hro 9 N² + 9h² + 18h n +3 9x²_3 lim hto gh² + 18h n h lim h (ght 180) h hro ght 18n 18x hto Therefore, f(x) - 180 giau