What is the slope of the line normal to the tangent line of $f(x) = e^-x+x^2-x$ at $x= 0$ ?

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Answer 1 · 2016-07-20T02:29:15

Slope of the normal to $f(x)$ at $x=0$ is $1/2$

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

Differentiating:
$f'(x) =-e^(-x)+2x-1$

To find the slope of $f(x)$ at $x=0$ :
$f'(0) = -1+0-1 =-2$

Hence slope of the tangent $(m)$ to $f(x)$ at $x=0$ is $-2$

Since the slope of the normal at the same point = $-1/m$
Therefore the slope of the normal to $f(x)$ at $x=0 = -1/(-2) = 1/2$

What is the slope of the line normal to the tangent line of f(x) = e^-x+x^2-x f(x) = e^-x+x^2-x at x= 0 x= 0 ?