Question

find the absolute extrema of the function on the closed interval.
y = 3x^{2/3}-2x, -1,1
minimum (x,y) = (
)
maximum (x,y) = (
)

Explanation:

Step1: Find the derivative of the function

The function is \(y = 3x^{\frac{2}{3}}-2x\).
Using the power rule \((x^n)^\prime=nx^{n - 1}\), we have \(y^\prime=3\times\frac{2}{3}x^{\frac{2}{3}-1}-2\).
Simplify to get \(y^\prime = 2x^{-\frac{1}{3}}-2=\frac{2}{x^{\frac{1}{3}}}-2=\frac{2 - 2x^{\frac{1}{3}}}{x^{\frac{1}{3}}}\).

Step2: Find the critical points

Set \(y^\prime = 0\), then \(\frac{2-2x^{\frac{1}{3}}}{x^{\frac{1}{3}}}=0\).
The numerator \(2 - 2x^{\frac{1}{3}}=0\) gives \(x^{\frac{1}{3}} = 1\), so \(x = 1\).
Also, \(y^\prime\) is undefined when \(x = 0\) (since the denominator \(x^{\frac{1}{3}}=0\) at \(x = 0\)).

Step3: Evaluate the function at the critical points and endpoints

• When \(x=-1\): \(y=3(-1)^{\frac{2}{3}}-2(-1)=3 + 2=5\).
• When \(x = 0\): \(y=3(0)^{\frac{2}{3}}-2(0)=0\).
• When \(x = 1\): \(y=3(1)^{\frac{2}{3}}-2(1)=3-2 = 1\).

Answer:

minimum \((x,y)=(0,0)\)
maximum \((x,y)=(-1,5)\)