What is the equation of the normal line of f(x)=(1+2x)⋅ex2 at x=−12?

The tangent line has equation&nbsp;y−y0=f′(x0)(x−x0)&nbsp;and normal liney−y0=−1f′(x0)(x−x0)Then,&nbsp;f'(x)=2ex2+(1+2x)ex2*2x==2ex2+2xex2+4x2ex2f'(-12)=2e4-e4+e4=2e4Normal line equation is&nbsp;y=-12e4(x+12)Because&nbsp;y0=0

Key to "What is the equation of the normal line..."

High School
Calculus and Analysis
Calculus 1
Derivatives

Trent Fuller

Answered question

2022-04-09

What is the equation of the normal line of

f (x) = (1 + 2 x) \cdot e^{x^{2}}

x = - \frac{1}{2}

Answer & Explanation

uvredio0of6

Beginner2022-04-10Added 17 answers

The tangent line has equation $y - y_{0} = f^{'} (x_{0}) (x - x_{0})$ and normal line
$y - y_{0} = - \frac{1}{f^{'} (x_{0})} (x - x_{0})$
Then, $f' (x) = 2 e^{x^{2}} + (1 + 2 x) e^{x^{2}} * 2 x =$
$= 2 e^{x^{2}} + 2 x e^{x^{2}} + 4 x^{2} e^{x^{2}}$
$f' (- \frac{1}{2}) = 2 \sqrt[4]{e} - \sqrt[4]{e} + \sqrt[4]{e} = 2 \sqrt[4]{e}$
Normal line equation is $y = - \frac{1}{2 \sqrt[4]{e}} (x + \frac{1}{2})$
Because $y_{0} = 0$

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