What is the instantaneous rate of change of #f(x)=ln(2x^2-4x+6) # at #x=0 #?

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Answer 1 · 2016-03-04T17:09:15

#-2/3#

#f(x)=ln(2x^2-4x+6)#

Using the chain rule:

#f'(x)=1/(2x^2-4x+6)xx(4x-4)#

This simplifies to:

#f'(x)=(2x-2)/(x^2-2x+3)#

When #x=0# this becomes:

#(0-2)/(0+0+3)=-2/3#

This is the instantaneous rate of change of the function at #x=0#