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

Apr 8, 2017

$y = x$

#### Explanation:

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

1. Point on the line: $\left({x}_{1} , {y}_{1}\right) = \left(1 , f \left(1\right)\right) = \left(1 , 1\right)$
2. Slope of the line: $m = - \frac{1}{f ' \left(1\right)} = - \frac{1}{- 1} = 1$
(Note: Since the normal line is perpendicular to the tangent line, we take the negative reciprocal of the slope of the tangent line $f ' \left(1\right)$.)

By Point-Slope Form $y - {y}_{1} = m \left(x - {x}_{1}\right)$,

$y - 1 = 1 \left(x - 1\right) R i g h t a r r o w y = x$

I hope that this was clear.

Here are more details:

$f \left(1\right) = \frac{1}{{e}^{2 \left(1\right) - 2}} = \frac{1}{e} ^ 0 = 1$

$f ' \left(x\right) = \frac{1 \cdot {e}^{2 x - 2} - x \cdot 2 {e}^{2 x - 2}}{{e}^{2 x - 2}} ^ 2 = \frac{\left(1 - 2 x\right) \cancel{{e}^{2 x - 2}}}{{e}^{2 x - 2}} ^ \left(\cancel{2}\right) = \frac{1 - 2 x}{e} ^ \left(2 x - 2\right)$

So,

$f ' \left(1\right) = \frac{1 - 2 \left(1\right)}{{e}^{2 \left(1\right) - 2}} = \frac{- 1}{e} ^ 0 = - 1$