How do you find the critical numbers for $f(x)= (x-1) e^x$ to determine the maximum and minimum?

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Answer 1 · 2017-02-28T17:34:41

The only critical point is for $x=0$ and the function has a local minimum in that point.

Critical points are found by equating the first derivative to zero:

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

Solving:

$xe^x = 0$

the only critical point is $x=0$

Now consider that:

$f'(x) = xe^x < 0$ for $x<0$

$f'(x) = xe^x > 0$ for $x>0$

We have then that $f(x)$ is decreasing in $(-oo,0)$ and increasing in $(0,+oo)$ , therefore $x=0$ is a local minimum.

graph{(x-1)e^x [-10, 10, -5, 5]}

How do you find the critical numbers for f(x)= (x-1) e^xf(x)= (x-1) e^x to determine the maximum and minimum?