Question

What is the instantaneous rate of change of `f(x)=ln(4x^2-2x )` at `x=-2`?

Answer 1

Instantaneous rate of change of `f(x)=ln(4x^2-2x)` at `x=-2` is `-9/10`

Explanation:

Instantaneous rate of change of a function is given by the value of the first derivative at that point.

As the function is `f(x)=ln(4x^2-2x)`

`f'(x)=1/(4x^2-2x)xx(8x-2)=(8x-2)/(4x^2-2x)`

and at `x=-2`, we have `f'(-2)=(8(-2)-2)/(4(-2)^2-2(-2))`

= `(-16-2)/(16+4)=-18/20=-9/10`