What is the equation of the normal line of f(x)=x^3e^-x+x^3 at x=-5?

Feb 12, 2016

${y}_{t} = {\textcolor{red}{m}}_{t} x + {\textcolor{red}{b}}_{t}$
The normal equation is ${y}_{n} = {\textcolor{b l u e}{m}}_{n} x + {\textcolor{b l u e}{b}}_{n}$
a) Where ${\textcolor{red}{m}}_{t} = - \frac{1}{\textcolor{b l u e}{m}} _ n$
b) ${\textcolor{red}{b}}_{t} = {\textcolor{b l u e}{b}}_{n}$

Explanation:

the slope of the normal line is the negative reciprocal of the slope of the tangent line. so what you need to don is :
1) find the the slope of f(x) at $x = - 5 \implies f ' \left(- 5\right)$
a) $f ' \left(x\right) = 3 {x}^{2} {e}^{-} x - {x}^{3} {e}^{-} x + 3 {x}^{2}$
b) $f ' \left(- 5\right) = y = 75 {e}^{5} + 125 {e}^{5} - 125$
2) find the equation of line, equation of tangent at (-5, f(5))
a) Tangent equation, ${y}_{t} = {\textcolor{red}{m}}_{t} x + {\textcolor{red}{b}}_{t}$
b) ${\textcolor{red}{m}}_{t} = f ' \left(- 5\right)$
c) ${y}_{t} - f \left(- 5\right) = \textcolor{red}{f} ' \left(- 5\right) \left(x + 5\right) + {\textcolor{red}{b}}_{t}$ solve for b
4) The normal equation is ${y}_{n} = {\textcolor{b l u e}{m}}_{n} x + {\textcolor{b l u e}{b}}_{n}$
a) Where ${\textcolor{red}{m}}_{t} = - \frac{1}{\textcolor{b l u e}{m}} _ n$
b) ${\textcolor{red}{b}}_{t} = {\textcolor{b l u e}{b}}_{n}$

There is some algebra that you need to, which I leave to you. the recipe for solving is there for you for now and future problems...

Good luck