I need to find for which natural values of $a$ the function $f(x)=x^ae^x$ has exactly one extremum. I calculated the derivative: $f'(x)=e^xx^{a-1}(a+x)$, but I don't know what to do. I know that I can't say that $f$ has only one extremum when the derivative has only one root, because it's not true.
How can I solve this?
On
Since $a$ is restricted to natural numbers, the derivative:
$$f'(x)_a=e^xx^{a-1}(a+x)$$
If we plug in the smallest natural number:
$$f'(x)_1=e^x(1+x)$$
Going up the number line,
$$f'(x)_2=e^xx(2+x)$$
$$f'(x)_3=e^xx^2(x+3)$$
$$f'(x)_4=e^xx^3(x+4)$$
$$f'(x)_5=e^xx^4(x+5)$$
From doing so, one can notice how the # of points of extremum will keep varying between 1 and 2 as $a$ varies from odd to even. (See sign change of derivative to make this observation)
As the clear pattern may suggest, there are no more points of extremum possible other than the roots of the derivative (some of them at least).
Thus, its clear how the values of $a$ where $f(x)$ has one extremum are all odd numbers.
I don't know why you say "it's not true". By the usual definition, a stationary point of $f$ is a zero of $f′$, so you are indeed looking for when $f'$ has one zero.
If $a>1$, both $x=0$ and $x=−a$ are stationary points. What happens if $a=0$? If $a=1$?