Find the local maxima and minima .

Question

Consider the function $ \large f(x)=-x^{\frac{2}{3}} (x-4) \ $ in domain $ \ [-4,4] \ $. Find the local maxima and minima .what is absolute maximum-minimum?

Answer:

I have found the only critical point $ x=\frac{8}{5} \ $

Using 1st-derivative test , I see that $ x=\frac{8}{5} \ $ is a local maxima.

So there is no minima.

Am I right?

Answer 1

The domain is $x>0$.

$$f'(x)=-\frac{5}{3}x^{\frac{2}{3}}+\frac{8}{3}x^{-\frac{1}{3}}=\frac{8-5x}{3x^{\frac{1}{3}}}.$$

We see that $f$ increases on $\left(0,\frac{8}{5}\right]$ and $f$ decreases on $\left[\frac{8}{5},+\infty\right)$.

Thus, $x_{max}=\frac{8}{5}$ and $f$ has no a local minimum point.

If you mean that $f(x)=-\sqrt[3]{x^2}(x-4)$ then the domain with our given is $[-4,4]$, $$f'(x)=\frac{8-5x}{3\sqrt[3]{x}}$$ which gives that $x_{min}=0$

because $f$ decreases on $[-4,0]$ and $f$ increases on $\left[0,\frac{8}{5}\right]$.