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

1 Answer

Shwetank Mauria

Jul 29, 2016

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#

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