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I just wanted to ask, why does the following function:

$$f(x)=x^{1/3}(x+3)^{2/3}$$

Have 3 crtical points $0,-1,-3$, because its first derivative is:

$$f'(x)=\frac{x+1}{x^{2/3}(x+3)^{1/3}}$$

$$f'(x)=\frac{x+1}{x^{2/3}(x+3)^{1/3}}=0$$

Then $x=-1$ is the critical point as its undefined on $-3$, and $0$, so why is $-3$, and $0$ also considered a critical point$?$

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  • $\begingroup$ @caverac Why need $x \ge 0$? Those are cuberoots in the formula. $\endgroup$
    – coffeemath
    Commented Nov 25, 2018 at 15:01

2 Answers 2

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A critical point $c$ is where $f(c)$ defined, and $f'(c)$ either zero or undefined.

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Hint critical points any limit points where the function may be prolongated by continuity or where the derivative is not defined. or a critical point or stationary point of a differentiable function of a real or complex variable is any value in its domain where its derivative is 0.

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    $\begingroup$ Note that the function needs to be defined at the critical point. No "prolongation by continuity" is used in the definition. $\endgroup$
    – coffeemath
    Commented Nov 25, 2018 at 14:58
  • $\begingroup$ some authors mentioned this point $\endgroup$
    – Unknown
    Commented Nov 25, 2018 at 15:00
  • $\begingroup$ John-- I'd say if one prolongs a domain by conytinuity one is working with a different function. $\endgroup$
    – coffeemath
    Commented Nov 25, 2018 at 15:03

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