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

1 Answer
Steve M · Jim H
Oct 11, 2016

the gradient of the normal at #x=1# is #m=-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'(1)# and hence we need to find #f'(x)#

#f(x)=e^(x^2-1)+3x-2#
#:. f'(x)=e^(x^2-1)(2x)+3 #

We don't need to simplify any more - just substitute #x=1#:
# x=1 => f'(1)=e^(1-1)(2)(1)+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=-1/5#

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