# What is the equation of the line that is normal to #f(x)= x/e^(2x-2) # at # x= 1 #?

To find an equation of a line, we need two things:

- Point on the line:
#(x_1,y_1)=(1,f(1))=(1,1)# - Slope of the line:
#m=-1/(f'(1))=-1/(-1)=1#

(Note: Since the normal line is perpendicular to the tangent line, we take the negative reciprocal of the slope of the tangent line#f'(1)# .)By Point-Slope Form

#y-y_1=m(x-x_1)# ,#y-1=1(x-1) Rightarrow y=x# I hope that this was clear.

Here are more details:

#f(1)=1/(e^(2(1)-2))=1/e^0=1# By Quotient Rule,

#f'(x)=(1cdote^(2x-2)-x cdot 2e^(2x-2))/(e^(2x-2))^2=((1-2x)cancel(e^(2x-2)))/(e^(2x-2))^(cancel(2))=(1-2x)/e^(2x-2)# So,

#f'(1)=(1-2(1))/(e^(2(1)-2))=(-1)/e^0=-1#

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To find the equation of the line that is normal to the function f(x) = x/e^(2x-2) at x = 1, we need to determine the slope of the tangent line at x = 1 and then find the negative reciprocal of that slope to obtain the slope of the normal line.

First, we find the derivative of f(x) with respect to x, which is f'(x) = (e^(2x-2) - 2x + 2) / e^(2x-2).

Next, we evaluate f'(1) to find the slope of the tangent line at x = 1. Plugging in x = 1 into f'(x), we get f'(1) = (e^(0) - 2 + 2) / e^(0) = 0.

Since the slope of the tangent line is 0, the slope of the normal line will be the negative reciprocal of 0, which is undefined.

Therefore, the equation of the line that is normal to f(x) = x/e^(2x-2) at x = 1 is undefined.

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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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