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

#y=-1/82.34x+22.09#

Given

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

At

#y=2.71828^(2^2-1)+2(2)-2#

#y=2.71828^3+4-2#

#y=20.085+2=22.085#

At Point

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

#dy/dx=2(2)(2.71828^3)+2#

#dy/dx=4xx 20.085+2#

#dy/dx=82.34#

The slope of the tangent

The slope of the normal is

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#

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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 ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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