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 $x=0$

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