What is the instantaneous rate of change of $f(x)=xe^x-(x-2)e^(3x-2)$ at $x=-1$ ?

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Answer 1 · 2016-01-17T06:29:13

$f' (-1)=8*e^-5$
$f' (-1)=0.053903575992684$

Find the first derivative $f' (x)$ then evaluate $f' (-1)$
It goes like this;

$f(x) = x e^x-(x-2) e^(3x-2)$

$f' (x) =x e^x*1+e^x*1-[(x-2)*e^(3x-2)*3+e^(3x-2) *1]$
now use $x=-1$

$f '(-1)=$
$(-1)e^-1+e^-1-[(-1-2) e^(3(-1)-2) * 3+e^(3(-1)-2)]$

$f' (-1)=0-(-3 e^-5 *3+e^-5)$

$f' (-1)=9e^-5-e^-5$

$f' (-1)=8 e^-5$

$f' (-1)= 0.053903575992669$

What is the instantaneous rate of change of f(x)=xe^x-(x-2)e^(3x-2) f(x)=xe^x-(x-2)e^(3x-2) at x=-1 x=-1 ?