# What is the slope of the line normal to the tangent line of f(x) = e^(x^2-1)+3x-2  at  x= 1 ?

Oct 11, 2016

the gradient of the normal at $x = 1$ is $m = - \frac{1}{5}$

#### Explanation:

To answer this question we need to find the slope of the tangent when $x = 1$; and use the fact that the normal and tangent are perpendicular the so the product of their slopes = -1

In order to this we need to evaluate $f ' \left(1\right)$ and hence we need to find $f ' \left(x\right)$

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

We don't need to simplify any more - just substitute $x = 1$:
$x = 1 \implies f ' \left(1\right) = {e}^{1 - 1} \left(2\right) \left(1\right) + 3 = 5$

So the gradient of the tangent at $x = 1$ is $m = 5$
So the gradient of the normal at $x = 1$ is $m = - \frac{1}{5}$