What is the slope of the line normal to the tangent line of f(x) = xe^-x  at  x= 0 ?

Jan 13, 2017

${m}_{n} = - \frac{1}{f ' \left({x}_{0}\right)} = - 1$

Explanation:

The slope of the tangent to the graph of $y = f \left(x\right)$ for $x = {x}_{0}$ is ${m}_{t} = f ' \left({x}_{0}\right)$.

So the slope of the normal to the tangent is:

${m}_{n} = - \frac{1}{m} _ t = - \frac{1}{f ' \left({x}_{0}\right)}$

In our case;

$f ' \left(x\right) = {e}^{- x} - x {e}^{- x} = {e}^{- x} \left(1 - x\right)$

For $x = 0$ the slope of the tangent is:

${m}_{n} = - \frac{1}{f ' \left({x}_{0}\right)} = - 1$

The equation of the normal line is:

$y = f \left({x}_{0}\right) - \frac{1}{f ' \left({x}_{0}\right)} \left(x - {x}_{0}\right)$

that is:

$y = - x$