Question

What is the equation of the line that is normal to `f(x)=(2x-4)^2/e^x` at `x=2`?

Answer 1

The vertical line `x=2`

Explanation:

The normal line passes through the point `(2,f(2))`. Note that `f(2)=(4-4)^2/e^2=0`, to the point is `(2,0)`.

The normal line is perpendicular to the tangent line. So, using the derivative, we can find the slope of the tangent line at `x=2`, then take its opposite reciprocal to find the slope of the normal line.

To find this derivative, let's use the quotient rule:

`f'(x)=((d/dx(2x-4)^2)e^x-(2x-4)^2(d/dxe^x))/(e^x)^2`

We'll need the chain rule for `d/dx(2x-4)^2`:

`f'(x)=(2(2x-4)^1(d/dx(2x-4))e^x-(2x-4)^2e^x)/e^(2x)`

Note that `d/dx(2x-4)=2`:

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

We can simplify, if we want:

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

`f'(x)=((2x-4)(8-2x))/e^x`

`f'(x)=(4(x-2)(3-x))/e^x`

So we see that `f'(2)=0`. This presents a tiny problem: we can't take the opposite reciprocal of `0`. But, a slope of `0` means that the tangent line is horizontal at `x=2`. Thus, the perpendicular normal line will be a vertical line. The vertical line that passes through the point `(2,0)` is the line `x=2`.