Locate the absolute extrema of the function on the closed interval:
y = 3x^(2/3) - 2x, [-1, 1]

Question

Locate the absolute extrema of the function on the closed interval:
y = 3x^(2/3) - 2x, [-1, 1]

anonymous · Answer

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

anonymous · Answer

^-- finding critical points, btw

anonymous · Answer

$$1-x^{1/3} = 0$$
$$x^{1/3} = 1$$
$$x = 1$$?? my book reads "0" as the critical point.

anonymous · Answer

0 is a critical point because that is where the derivative is undefined.
the definition of critical point is where f' = 0 OR f' does not exist.

anonymous · Answer

i understand, but by setting the derivative = to 0 and solving for x, i've defined the critical point for which the derivative of the function is equal to zero, no? as shown in my work above

anonymous · Answer

yes... so in essence, in the interval [-1, 1], you have two critical numbers.

anonymous · Answer

ummm wait... are you looking at [-1, 1] or (-1, 1)  ???

anonymous · Answer

in the closed interval [-1,1]