Critical points of a function and discontinuity

Question

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$?$

@caverac Why need $x \ge 0$? Those are cuberoots in the formula. — coffeemath, Nov 25 '18 at 15:01

score 2 · Accepted Answer · answered Nov 25 '18 at 14:57

2

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

answered Nov 25 '18 at 14:57

coffeemath

7,403

Unknown · Answer 2 · 2018-11-25T14:59:57.477

0

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.

edited Nov 25 '18 at 14:59

answered Nov 25 '18 at 14:56

Unknown

3,073

1

Note that the function needs to be defined at the critical point. No "prolongation by continuity" is used in the definition. – coffeemath Nov 25 '18 at 14:58
some authors mentioned this point – Unknown Nov 25 '18 at 15:00
John-- I'd say if one prolongs a domain by conytinuity one is working with a different function. – coffeemath Nov 25 '18 at 15:03

Critical points of a function and discontinuity

2 Answers2