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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.See more: Left To Tell Sparknotes - Left To Tell Summary And Study Guide**

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