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

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Answer 1 · 2017-01-13T15:00:38

$m_n = -1/(f'(x_0)) = -1$

The slope of the tangent to the graph of $y=f(x)$ for $x=x_0$ is $m_t = f'(x_0)$ .

So the slope of the normal to the tangent is:

$m_n =-1/m_t = -1/(f'(x_0))$

In our case;

$f'(x) = e^(-x) -xe^(-x) = e^(-x) (1-x)$

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

$m_n = -1/(f'(x_0)) = -1$

The equation of the normal line is:

$y= f(x_0) -1/(f'(x_0))(x-x_0)$

that is:

$y = -x$

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