# Find the local maxima and minima .

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?

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?

• According to Desmos, you're right! Notice that after the critical point the graph goes down and down, and that the function is not defined at $x=0$. – Toby Mak Oct 23 '17 at 12:03
• The function is defined at $x=0 \$ – M. A. SARKAR Oct 23 '17 at 12:10

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

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

• But in domain $[-4,4] \$ , does $x=0 \$ gives local minima ? – M. A. SARKAR Oct 23 '17 at 12:06
• @mabmath No! If we write $x^{\frac{1}{3}}$ then the domain is $x>0$. Otherwise, if we write $\sqrt{x}$ then the domain is $\mathbb R$. – Michael Rozenberg Oct 23 '17 at 12:08
• The function in fact has domain $\mathbb{R}$. Only its derivative is not continuous at $0$ – ehochix Oct 23 '17 at 12:09
• @Cuteboy If $f(x)=-\sqrt{x^2}(x-4)$ then the domain of $f$ is indeed, $\mathbb R$. The domain of $f(x)=-x^{\frac{2}{3}}(x-4)$ is $(0,+\infty)$. – Michael Rozenberg Oct 23 '17 at 12:11
• @MichaelRozenberg I think $x^{\frac{2}{3}}$ is understood as $(x^2)^{\frac{1}{3}}$, so it is well defined for all real $x$. – ehochix Oct 23 '17 at 12:12