# What is the equation of the normal line of f(x)=(1-x)e^(3x)-e^x at x=1?

Nov 8, 2016

graph{((1-x)e^(3x)-e^x-y)(y+e-1/(e(e^2+1))(x-1))=0 [-3.844, 4.926, -3.28, 1.103]}

$y + e = \frac{1}{e \left({e}^{2} + 1\right)} \left(x - 1\right)$

#### Explanation:

The general equation for normal is

$y - f \left({x}_{0}\right) = - \frac{1}{f} ^ ' \left({x}_{0}\right) \left(x - {x}_{0}\right)$

${x}_{0} = 1$

$f \left({x}_{0}\right) = f \left(1\right) = - e$

${f}^{'} \left(x\right) = - {e}^{x} \left(3 x {e}^{2 x} + 1 - 2 {e}^{2 x}\right)$

${f}^{'} \left({x}_{0}\right) = {f}^{'} \left(1\right) = - e \left(3 {e}^{2} + 1 - 2 {e}^{2}\right) = - e \left({e}^{2} + 1\right)$

$y + e = \frac{1}{e \left({e}^{2} + 1\right)} \left(x - 1\right)$