## University Calculus: Early Transcendentals (3rd Edition)

At $x=e^{-1}$, $y$ has the absolute and local minimum value of $-e^{-1}$. There is no absolute or local maximum value of $y$.
$$y=x\ln x$$ - Domain: $(0,\infty)$ 1) Find all the critical points of the function: - Find $y'$: $$y'=(x)'\ln x+x(\ln x)'=\ln x+x\times\frac{1}{x}$$ $$y'=\ln x+1$$ - We have $y'=0$ when $$\ln x=-1$$ $$x=e^{-1}$$ - There is no value of $x$ in the defined domain $(0,\infty)$ for which $y'$ is not defined. So function $y$ has one critical point $x=e^{-1}$. 2) Evaluate $y$ at the critical point: - For $x=e^{-1}$: $$y=e^{-1}(\ln e^{-1})=e^{-1}(-1)=-e^{-1}$$ Now take a look at the graph of $y$. From the graph, we see that at $x=e^{-1}$, $y$ has the absolute and local minimum value of $-e^{-1}$. There is no absolute or local maximum value of $y$ though.