Question #8a726

1 Answer
Somebody N.
Jan 8, 2018

See below.

Explanation:

To find the instantaneous rate of change, we need to find the derivative of #1/x#. Putting in values of #x# will give us the gradient of the tangent at that point. This is different from the average rate of change, which is the gradient of the secant line that joins #2# points.

So the instantaneous rate of change is given by #f'(a)#, and the average rate of change on an interval #[a,b]# is given by

#((f(b))-(f(a)))/(b-a)#

#1/x=x^-1#

Using the power rule:

#dy/dx(x^n)=nx^(n-1)#

#dy/dx(x^-1)=-x^(-2)=-1/x^2#

#:.#

#x_0=2#

#f'(x_0)=-1/(2)^2=color(blue)(-1/4)#

#x_1=3#

#f'(x_1)=-1/(3)^2=color(blue)(-1/9)#

Instantaneous Rate of Change:

enter image source here

Average Rate of Change:

enter image source here

Related questions
See all questions in Instantaneous Rate of Change at a Point
Impact of this question
1433 views around the world
You can reuse this answer
Creative Commons License