Respuesta :

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:

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

Now,

→ [tex]\frac{dy}{dx} = 2x^{-\frac{1}{3} } -2[/tex]

       [tex]= \frac{2}{x^{\frac{1}{3} }} -2[/tex]

       [tex]= \frac{2[1-x^{\frac{1}{3} }]}{x^{\frac{1}{3} }}[/tex]

Critical point,

  • x = 0

By evaluating the function values at the critical point,

[tex]x =0[/tex]

→ [tex]f(0) =0[/tex]

[tex]x =-1[/tex]

→ [tex]f(-1) = 3+2[/tex]

             [tex]= 5[/tex]

[tex]x =1[/tex]

→ [tex]f(1) = 3-2[/tex]

          [tex]= 1[/tex]

Thus the above answer is correct.

Learn more about function here:

https://brainly.com/question/1931564

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