Answer
Function $y$ has the absolute and local minimum value of $-4$ at $x=4$ and the local maximum value of $0$ at $x=0$. There is no absolute maximum value.
Work Step by Step
$$y=x-4\sqrt x=x-4x^{1/2}$$
- Domain: $[0,\infty)$
1) Find all the critical points:
Find the derivative of the function: $$y'=1-4\times\frac{1}{2}x^{-1/2}=1-2x^{-1/2}$$ $$y'=1-\frac{2}{\sqrt x}$$
- For $y'=0$: $$1-\frac{2}{\sqrt x}=0$$ $$\frac{2}{\sqrt x}=1$$ $$\sqrt x=2$$ $$x=4$$
- For $y'$ to be undefined, $x=0$.
So $x=0$ and $x=4$ are the critical points of function $y$.
2) Evaluate function $y$ at the critical points:
- For $x=0$: $$y=0-4\sqrt0=0$$
- For $x=4$: $$y=4-4\sqrt4=4-4\times2=4-8=-4$$
Now check the graph of function $y=x-4\sqrt x$.
We see that function $y$ has the absolute and local minimum value of $-4$ at $x=4$ and the local maximum value of $0$ at $x=0$.
There is no absolute maximum value since as $x\to\infty$, $y\to\infty$ as well.
