How do you find the average and instantaneous rate of change of y=-1/(x-3)y=−1x−3 over the interval [0,1/2]?

1 Answer
Binayaka C.
Jul 9, 2018

Average rate change over the interval [0,1/2][0,12] is 2/15215
Instantaneous rate of change at point xx is 1/(x-3)^2 1(x−3)2

Explanation:

y=f(x)= -1/(x-3)= -(x-3)^-1y=f(x)=−1x−3=−(x−3)−1

Average rate of change over the interval [0,1/2][0,12] is

R_a= (f(b)- f(a))/(b-a) ;a= 0 , b= 1/2Ra=f(b)−f(a)b−a;a=0,b=12

f(1/2)= -1/(1/2-3)= 2/5 ; f(0)= -1/(0-3)= 1/3f(12)=−112−3=25;f(0)=−10−3=13

:. R_a= (2/5-1/3)/(1/2-0)=1/15/1/2=2/15

Average rate change over the interval [0,1/2] is 2/15

Instantaneous rate of change is derivative of f(x) at the

point x :. f^'(x)=-.-(x-3)^-2= 1/(x-3)^2

Instantaneous rate of change at point x is 1/(x-3)^2 [Ans]

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