What are the local extrema of $f (x) = x e^{- x}$ $f(x)= xe^-x$ ?

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Answer 1 · 2016-10-19T22:25:44

$(1, e^{- 1})$ $(1,e^-1)$

We need to use the product rule: $\frac{d}{d x} (u v) = u \frac{d v}{d x} + v \frac{d u}{d x}$ $d/dx(uv) = u(dv)/dx+v(du)/dx$

$:. f'(x) = xd/dx(e^-x) + e^-x d/dx(x)$
$:. f'(x) = x(-e^-x) + e^-x (1)$
$:. f'(x) = e^-x-xe^-x$

At a min/max $f'(x)=0$
$f'(x)=0 => e^-x(1-x) = 0$
Now, $e^x > 0 AA x in RR$
$:. f'(x)=0 => (1-x) = 0 => x=1$

$x=1 => f(1)=1e^-1 = e^-1$

Hence, there a single turning point at $(1,e^-1)$

graph{xe^-x [-10, 10, -5, 5]}

What are the local extrema of f(x)= xe^-xf(x)=xe−xf(x)= xe^-x?