Question

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

Answer 1

\[y' = 2*\frac{(1-x^{1/3})}{x^{1/3}}\]
my book reads "0" as a critical number; however, after solving for both fractions = 0, i get:
\[1 - x^{1/3} = 0\]
\[x = 1\]

Answer 2

1 should be the critical number, not zero, no?

Answer 3

It looks like they both are critical values.

Answer 4

can you explain how you found 0 as the critical value.

Answer 5

\[\frac{2}{x^{1/3}} = 0\]
\[2 = 0\]
... this is the second fraction.