Question

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

Answer 1

`2y+x=1`

Explanation:

`f(x)= (2x-2)e^(2x-2)`
`implies f'(x) = (4x-2)*e^(2x-2)`

`implies f'(1) = 2`
`implies (dy/dx)_"at x =1" = 2`

`implies (dx/dy)_"at x =1" = -1/2`

Hence we get the slope of the normal: `m=-1/2`

Also, `f(1)=0`, hnce f(x) passes through `(1,0)`
Using point slope form of line,

`(y-0)=m(x-1)`
`implies y=-1/2(x-1)`
`implies 2y+x=1`