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Verified Answer :

Kmidgetchknfighter15

y'=2x^-1/3-2=0, u get x^1/3=1, so x=1, then y=1, but interval [-1,1], we have 3 values of x:-1,0,1 
x=1 so y=1, x=0 so y=0, x=-1 so y=5 To get the extrema, derive the function. You get y' = 2x^-1/3 - 2. Set this equal to zero, and you get x=0 as the location of a critical point. Since you are on a closed interval [-1, 1], those points can also have an extrema

X=1  min at (0,0) max at (-1,5)

Cricetus

The absolute minimum is "(0, 0)" and absolute maximum is "(-1, 5)".

Given:

  • \(y = 3x^{\frac{2}{3} }-2x\)
  • \(x \ \epsilon \ [-1, 1]\)

Now,

→ \(\frac{dy}{dx} = 2x^{-\frac{1}{3} } -2\)

       \(= \frac{2}{x^{\frac{1}{3} }} -2\)

       \(= \frac{2[1-x^{\frac{1}{3} }]}{x^{\frac{1}{3} }}\)

Critical point,

  • x = 0

By evaluating the function values at the critical point,

\(x =0\)

→ \(f(0) =0\)

\(x =-1\)

→ \(f(-1) = 3+2\)

             \(= 5\)

\(x =1\)

→ \(f(1) = 3-2\)

          \(= 1\)

Thus the above answer is correct.

Learn more about function here:

https://brainly.com/question/1931564

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