What are the critical points of f(x) =xsqrt(e^x-3x)?

Feb 8, 2016

They are approximately $0.619$, $1.512$, and $0.395$.

Explanation:

f'(x) = (xe^x-2e^x-9x)/(2sqrt(e^x-3x)

The critical numbers of $f$ are the solutions to

${e}^{x} - 3 x = 0$

and the solution to $x {e}^{x} - 2 {e}^{x} - 9 x = 0$ that gives ${e}^{x} - 3 x \ge 0$ (so that it is in the domain of $f$.

Use whatever numerical/technological methods you have to get approximations.

$0.619$ and $1.512$ solve the first equation and

$0.395$ and $1.234$ solve the second, but ${e}^{1.234} - 3 \left(1.234\right) < 0$ so $1.234$ is not in the domain of $f$ and hence is not a critical number for $f$.