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

Question

2170 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-26T20:26:46

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 $y_0=0$

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