So for this equation, I already found the derivative and now to find the critical points, I have to set it equal to 0 and solve for x.
The equation is: $f'(x) = 2x\ln x + x = 0$
Thanks for the help guys.
calculusderivatives
So for this equation, I already found the derivative and now to find the critical points, I have to set it equal to 0 and solve for x.
The equation is: $f'(x) = 2x\ln x + x = 0$
Thanks for the help guys.
Best Answer
Factor out the $x$: $$x(2 \ln x + 1) = 0 \implies x = 0 \text{ or } \ln x = -\frac 12$$
Now, what must $x$ be for $\;\ln x = -\frac 12\;$?
Hint: raise each side to a power of $e$: $$\ln x = -\frac 12 \iff e^{\ln x} = e^{-1/2} \quad \iff \quad x = e^{-1/2} = \frac{1}{\sqrt e}$$
So our two solutions to the equation are $x = 0,$ and $\;x = \dfrac{1}{\sqrt e}.$