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

Function $y$, at $x=0$, has the local minimum value of $1$.
$$y=\frac{1}{\sqrt[3]{1-x^2}}$$ $y$ is not defined where $\sqrt[3]{1-x^2}=0$, or $1-x^2=0$, or $x=\pm1$. So the domain of $y$ is $(-\infty,-1)\cup(-1,1)\cup(1,\infty)$. 1) Find all the critical points: Find the derivative of the function: $$y'=\frac{(1)'\sqrt[3]{1-x^2}-1(\sqrt[3]{1-x^2})'}{(\sqrt[3]{1-x^2})^2}$$ $$y'=\frac{0\times\sqrt[3]{1-x^2}-\Big((1-x^2)^{1/3}\Big)'}{(\sqrt[3]{1-x^2})^2}$$ $$y'=-\frac{\frac{1}{3}(1-x^2)^{-2/3}(1-x^2)'}{(\sqrt[3]{1-x^2})^2}$$ $$y'=-\frac{\frac{1}{3}(1-x^2)^{-2/3}(-2x)}{(\sqrt[3]{1-x^2})^2}$$ $$y'=\frac{\frac{2}{3}x(1-x^2)^{-2/3}}{(\sqrt[3]{1-x^2})^2}=\frac{\frac{2}{3}x(\sqrt[3]{1-x^2})^{-2}}{(\sqrt[3]{1-x^2})^2}$$ $$y'=\frac{2x}{3(\sqrt[3]{1-x^2})^4}$$ - For $y'=0$, $x=0$. - For $y'$ to be undefined, $\sqrt[3]{1-x^2}=0$, or $x^2=1$, or $x=\pm1$. However, since $x=\pm1$ does not lie in the defined domain, only $x=0$ is the critical point here. 2) Evaluate function $y$ at the critical points: - For $x=0$: $$y=\frac{1}{\sqrt[3]{1-0^2}}=\frac{1}{\sqrt[3]1}=1$$ Now check the graph of function $y$. We see that function $y$, at $x=0$, has the local minimum value of $1$. Function $y$ contains no absolute extrema since as $y\to\infty$ and $-\infty$ as $x$ approaches various points.