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

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Answer 1 · 2016-07-29T07:32:09

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

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 $f'(-2)=(8(-2)-2)/(4(-2)^2-2(-2))$

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

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