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

Question

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

Emily Ammerman · Accepted Answer

#y=-1/82.34x+22.09#

Given

#y=e^(x^2-1)+2x-2#

At #x=2#

#y=2.71828^(2^2-1)+2(2)-2#
#y=2.71828^3+4-2#
#y=20.085+2=22.085#

At Point #(2, 22.085)# There is a tangent to the curve.
Look at the graph

The slope of the tangent is equal to the slope of the given curve.

The slope of the given curve is-

#dy/dx=2xe^(x^2-1)+2#

At #x=2# the slope of the curve is

#dy/dx=2(2)(2.71828^3)+2#
#dy/dx=4xx 20.085+2#
#dy/dx=82.34#

The slope of the tangent #m_1=82.34#

The slope of the normal is #m_2=-1xx1/82.34=(-1)/82.34#

The equation of the Normal is -

#y=m_2x+c#
#22.085=-1/82.34(2)+c#
#22.085+1/82.34=c#
#22.09=c#
#y=-1/82.34x+22.09#

Scarlett Allison · Answer

undefined To find the slope of the line normal to the tangent line of $ f(x) = e^{(x^2-1)} + 2x - 2 $ at $ x = 1 $, we first find the derivative of $ f(x) $ and evaluate it at $ x = 1 $ to get the slope of the tangent line. Then, we find the negative reciprocal of this slope to get the slope of the line normal to the tangent line.

The derivative of $ f(x) $ is $ f'(x) = 2xe^{(x^2-1)} + 2 $.

Evaluating $ f'(1) $ gives $ f'(1) = 2e^{(1^2-1)} + 2 = 2e^0 + 2 = 2 + 2 = 4 $.

Therefore, the slope of the tangent line at $ x = 1 $ is $ m = 4 $.

The slope of the line normal to the tangent line is the negative reciprocal of the slope of the tangent line, so the slope of the line normal to the tangent line is $ -1/4 $.