Question

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

Answer 1

`y=x`

Explanation:

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

1. Point on the line: `(x_1,y_1)=(1,f(1))=(1,1)`
2. 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`