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

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Answer 1 · 2016-09-08T14:49:37

The reqd. slope $=-1/(e+1)$ .

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

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

As the normal to $C$ at $P$ is $bot$ to tgt., its slope is, $-1/(f'(a))$ ,

if, $f'(a) ne 0$ .

We have, $f(x)=e^(x-1)+x-2 rArr f'(x)=e^(x-1)+1$

$rArr f'(2)=e+1 ne0$ .

Therefore, the reqd. slope $=-1/(e+1)$ .

Enjoy Maths.!

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