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

The critical points are $(0,0), (2.4,4.462)$ and $(3,0)$. $(2.4,4.462)$ is also the endpoint of $y$ here. Function $y$ has the absolute and local minimum value of $0$ at $x=0$ and $x=3$, while it also has the local maximum value of $4.462$ at $x=2.4$. There is no absolute maximum value.
$$y=x^2\sqrt{3-x}$$ - Domain: $(-\infty,3]$ The domain shows that there exists an endpoint at $x=3$. 1) Find the derivative of function $y$: $$y'=(x^2)'\sqrt{3-x}+x^2(\sqrt{3-x})'$$ $$y'=2x\sqrt{3-x}+\frac{x^2(3-x)'}{2\sqrt{3-x}}$$ $$y'=2x\sqrt{3-x}+\frac{x^2(-1)}{2\sqrt{3-x}}$$ $$y'=2x\sqrt{3-x}-\frac{x^2}{2\sqrt{3-x}}$$ $$y'=\frac{4x(3-x)-x^2}{2\sqrt{3-x}}=\frac{12x-4x^2-x^2}{2\sqrt{3-x}}$$ $$y'=\frac{12x-5x^2}{2\sqrt{3-x}}$$ - We have $y'=0$ when $$12x-5x^2=0$$ $$x(12-5x)=0$$ $$x=0\hspace{1cm}\text{or}\hspace{1cm}x=\frac{12}{5}$$ - $y'$ is undefined when $2\sqrt{3-x}=0$, or $x=3$. Therefore, the critical points are the ones at $x=0$, $x=12/5$ and $x=3$. 2) Evaluate $y$ at all the critical points and endpoints: - For $x=0$: $$y=0^2\sqrt{3-0}=0$$ - For $x=12/5=2.4$: $$y=\Big(\frac{12}{5}\Big)^2\sqrt{3-\frac{12}{5}}=\frac{144}{25}\sqrt{\frac{3}{5}}\approx4.462$$ - For $x=3$: $$y=3^2\sqrt{3-3}=9\sqrt0=0$$ So the critical points are $(0,0), (2.4,4.462)$ and $(3,0)$. $(2.4,4.462)$ is also the endpoint of $y$ here. Now take a look at the graph of function $y$. We see that function $y$ has the absolute and local minimum value of $0$ at $x=0$ and $x=3$, while it also has the local maximum value of $4.462$ at $x=2.4$. There is no absolute maximum value.