find the absolute extrema of the function on the closed interval.\ny = 3x^{2/3}-2x, -1,1\nminimum (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))