# What is the instantaneous rate of change of f(x)=(x^2-3x)e^(x-1)  at x=0 ?

Aug 1, 2016

$- \frac{3}{e}$

#### Explanation:

The $\textcolor{b l u e}{\text{instantaneous rate of change}}$ is the value of f'(0)

differentiate f(x) using the $\textcolor{b l u e}{\text{product rule}}$

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{f \left(x\right) = g \left(x\right) h \left(x\right) \Rightarrow f ' \left(x\right) = g \left(x\right) h ' \left(x\right) + h \left(x\right) g ' \left(x\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

here $g \left(x\right) = \left({x}^{2} - 3 x\right) \Rightarrow g ' \left(x\right) = 2 x - 3$

and $h \left(x\right) = {e}^{x - 1} \Rightarrow h ' \left(x\right) = {e}^{x - 1} . \frac{d}{\mathrm{dx}} \left(x - 1\right) = {e}^{x - 1}$
$\textcolor{b l u e}{\text{----------------------------------------------------------------------}}$

$\Rightarrow f ' \left(x\right) = \left({x}^{2} - 3 x\right) {e}^{x - 1} + {e}^{x - 1} . \left(2 x - 3\right)$

$= {e}^{x - 1} \left({x}^{2} - 3 x + 2 x - 3\right) = {e}^{x - 1} \left({x}^{2} - x - 3\right)$
$\textcolor{b l u e}{\text{-------------------------------------------------------------------}}$

$\Rightarrow f ' \left(0\right) = {e}^{0 - 1} \left(0 - 0 - 3\right) = - 3 {e}^{-} 1 = - \frac{3}{e}$