What is the equation of the normal line of $f(x)=xe^x$ at $x=3$ ?

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Answer 1 · 2018-07-19T15:15:57

$y-3e^3=-1/{4e^3}(x-3)$

Substituting $x=3$ in the given function, we get y-coordinate of point is obtained as follows

$y=f(3)=3e^3$

Differentiating given function w.r.t. $x$ , the slope of tangent $dy/dx$ is

$dy/dx=f'(x)$

$=d/dx(xe^x)$

$=xe^x+e^x$

hence the slope of tangent at $x=3$ is

$=f'(3)$

$=3e^3+e^3$

$=4e^3$

hence the slope $m$ of normal at the same point $x=3$ is given as

$m=-1/{f'(3)}$

$=-1/{4e^3}$

Now, the equation of normal at the point $(3, 3e^3)$ & having slope $m=-1/{4e^3}$ is given as follows

$y-3e^3=-1/{4e^3}(x-3)$

What is the equation of the normal line of f(x)=xe^xf(x)=xe^x at x=3 x=3 ?