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.!