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

Sep 8, 2016

The reqd. slope$= - \frac{1}{e + 1}$.

#### Explanation:

We know that, for a curve $y = f \left(x\right) , f ' \left(a\right)$ gives the slope of tgt.

line to $C$ at the pt.$P \left(a , f \left(a\right)\right)$.

As the normal to $C$ at $P$ is $\bot$ to tgt., its slope is, $- \frac{1}{f ' \left(a\right)}$,

if, $f ' \left(a\right) \ne 0$.

We have, $f \left(x\right) = {e}^{x - 1} + x - 2 \Rightarrow f ' \left(x\right) = {e}^{x - 1} + 1$

$\Rightarrow f ' \left(2\right) = e + 1 \ne 0$.

Therefore, the reqd. slope$= - \frac{1}{e + 1}$.

Enjoy Maths.!