Question

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

Answer 1

The tangent line has equation `y-y_0=f´(x_0)(x-x_0)` and normal line `y-y_0=-1/(f´(x_0))(x-x_0)`

Then, `f´(x)=2e^(x^2)+(1+2x)e^(x^2)·2x=`

`=2e^(x^2)+2xe^(x^2)+4x^2e^(x^2)`

`f´(-1/2)=2root(4)e-root(4)e+root(4)e=2root(4)e`

Normal line equation is `y=-1/(2root(4)e)(x+1/2)`

Because `y_0=0`