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?

  • 1
    $\begingroup$ 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 $x=0$. $\endgroup$ – Toby Mak Oct 23 '17 at 12:03
  • $\begingroup$ The function is defined at $ x=0 \ $ $\endgroup$ – M. A. SARKAR Oct 23 '17 at 12:10

The domain is $x>0$.


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 $\left[0,\frac{8}{5}\right]$.

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

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