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

The reqd. slope

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

First, let's find the derivative of ( f(x) ): [ f'(x) = \frac{d}{dx} (e^{x-1} + x - 2) ]

Using the chain rule and derivative of ( e^x ), we have: [ f'(x) = e^{x-1} + 1 ]

Now, let's evaluate ( f'(x) ) at ( x = 2 ): [ f'(2) = e^{2-1} + 1 = e + 1 ]

This is the slope of the tangent line at ( x = 2 ).

The slope of the line normal to the tangent line is the negative reciprocal of the slope of the tangent line. Therefore, the slope of the line normal to the tangent line at ( x = 2 ) is ( -\frac{1}{e + 1} ).

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