$\begingroup$
For $x > 0$, the graph takes on a parabola-like shape.
Is there a method to algebraically and not graphically determine the vertex of the graph?
2 Answers
$\begingroup$Note that
$$y’=y(\ln x+1)=0$$ yields $x=e^{-1}$ and $y(e^{-1})= e^{-1/e}$. Thus, the vertex is $(e^{-1}, e^{-1/e})$.
$\endgroup$ $\begingroup$$$y=x^x$$$$\frac{dy}{dx}=x^x(1 + \log x)$$Here, base of logarithm is $e=2.718281828459045....$
By definition, vertex is the point at which slope of tangent is zero, i.e. $\frac{dy}{dx}$ is zero.
So, for $\frac{dy}{dx}=0$, we must have:$$x^x(1+\log x)=0$$Hence, either $x^x=0$ (which is impossible), or $1+\log x=0$. Hence $\log x = -1$, which gives us $x=\frac{1}{e}$ and $y=\frac{1}{e^{\frac{1}{e}}}$. So, the vertex is $(\frac{1}{e},\frac{1}{e^{\frac{1}{e}}})$. Ta-da!
$\endgroup$
