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

Question

1164 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-10T05:21:16

$x+2y-1=0$ . See normal-inclusive Socratic graph.

graph{((2x-2)/(e^(2x-2))-y)(x+2y-1)=0 [-10, 10, -5, 5]}

$f =2e^2e^(-2x)(x-1)=0$ , at x = 1.

So, the foot of the normal is P( 1, 0 ).

f'=2e^2(e^(-2x)(x-1)'+(x-1)(e^(-2x))')#

$=2e^2(e^(-2x)-2e^(-2x)(x-1))=2$ , at x = 1.

Slope of the normal $= -1/(f')=-1/2$ .

So, the equation to the normal is

$y-0=-1/2(x-1)$ , giving

$x+2y-1=0$

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