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

Question

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

Elijah Abram · Accepted Answer

the gradient of the normal at #x=1# is #m=-1/5# 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#

Scarlett Allison · Answer

undefined To find the slope of the line normal to the tangent line of $ f(x) = e^{x^2-1} + 3x - 2 $ at $ x = 1 $, first, find the derivative of $ f(x) $ using the chain rule. Then, evaluate the derivative at $ x = 1 $ to find the slope of the tangent line. The slope of the line normal to the tangent line will be the negative reciprocal of the slope of the tangent line.