Question

What is the equation of the line normal to `f(x)= x^3e^x` at `x=2`?

Answer 1

`y=-x/(20*e^2)+1/(10*e^2)+8*e^2`

Explanation:

From the given equation `f(x)=x^3*e^x` at `x=2`

Let us solve for the point first. We should find `f(2)`

`f(2)=2^3*e^2=8*e^2`
We now have the point at `(2, 8e^2)`

Now, we need to compute the slope m and then obtain its negative reciprocal because, we are after the normal line.

First derivative of `f(x)` is `f' (x)`

`d/dxf(x)=f' (x)=d/dx(x^3*e^x)=e^x*d/dx(x^3)+x^3*d/dx(e^x)`

`f' (x)=3x^2*e^x+x^3*e^x`

for slope `m=f' (2)` use `x=2`

`f' (x)=3x^2*e^x+x^3*e^x`
`f' (2)=3(2)^2*e^2+(2)^3*e^2`
`f' (2)=12*e^2+8*e^2`
`f' (2)=20*e^2`
and `m=f' (2)=20*e^2`

For the normal line, we need `m'=-1/m`

`m'=-1/m=-1/(20*e^2)`

Let us write the Normal line equation using `m'=-1/(20*e^2)` and the point `(2, 8e^2)`

By Point-Slope form

`y-y_1=m' (x-x_1)`

`y-8e^2=-1/(20*e^2)(x-2)`

`y=-1/(20*e^2)(x-2)+8e^2`

`color(red)(y=-x/(20*e^2)+1/(10*e^2)+8*e^2" ")`the required Normal Line

God bless....I hope the explanation is useful.