What is the equation of the line normal to f(x)= x^3/e^x  at x=2?

Jan 8, 2017

$y = \frac{8}{e} ^ 2 - {e}^{2} / 4 \left(x - 2\right)$

Explanation:

The equation of the line normal to the graph of the function:

$y = f \left(x\right)$

at ath point $\left(\overline{x} , f \left(\overline{x}\right)\right)$ is given by:

$y = f \left(\overline{x}\right) - \frac{1}{f ' \left(\overline{x}\right)} \left(x - \overline{x}\right)$

In our case:

$f \left(x\right) = {x}^{3} / {e}^{x} = {x}^{3} {e}^{- x}$

$f \left(2\right) = \frac{8}{e} ^ 2$

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

$f ' \left(2\right) = \frac{12}{e} ^ 2 - \frac{8}{e} ^ 2 = \frac{4}{e} ^ 2$

So the normal line is:

$y = \frac{8}{e} ^ 2 - {e}^{2} / 4 \left(x - 2\right)$