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You are watching: Find the absolute extrema of the function on the closed interval. y = 3x2/3 − 2x, [−1, 1]

Given: `y=3x^(2/3)-2x,<-1,1>`

Find the crucial values for x by setup the derivative equal to zero and also solving for the x value(s).

`y"=2x^(-1/3)-2=0`

`2/x^(1/3)=2`

`x^(1/3)=1`

`x=1`

Plug in the an important x value(s) and the endpoints the the close up door interval right into the y(x) function.

`y(x)=3x^(2/3)-2x`

`y(-1)=5`

`y(1)=1`

research the y(x) worths to determine...

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Given: `y=3x^(2/3)-2x,<-1,1>`

Find the critical values for x by setting the derivative same to zero and also solving because that the x value(s).

`y"=2x^(-1/3)-2=0`

`2/x^(1/3)=2`

`x^(1/3)=1`

`x=1`

Plug in the an important x value(s) and the endpoints the the closed interval right into the y(x) function.

`y(x)=3x^(2/3)-2x`

`y(-1)=5`

`y(1)=1`

Examine the y(x) worths to recognize the absolute extrema.

The absolute maximum is the allude (-1, 5).

The absolute minimum is the allude (1, 1).

You need to find out the absolute extrema that the offered function, hence, you require to distinguish the role with respect come x, such that:

`y" = 3*(2/3)*x^(2/3 - 1) - 2`

You must solve for x the equation y" = 0

`3*(2/3)*x^(2/3 - 1) - 2 = 0`

`2/(x^(1/3)) - 2 = 0`

Factoring the end 2 yields:

`2(1/(root(3)x) - 1) = 0 => 1/(root(3)x) - 1 = 0`

`1 - root(3)x = 0 => root(3)x = 1 => x = 1^3 => x = 1 in <-1,1>`

Hence, examining the pure extrema that the provided function, over the term <-1,1>, yields that that reaches it"s extrema in ~ x = 1.

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