# `g(t) = (t^2)/(t^2 + 3), [-1,1]` Find the absolute extrema of the function on the closed interval. Given: `g(t)=(t^2)/(t^2+3)[-1,1]`

Find the critical values for t by setting the derivative equal to zero and solving for the t value(s).

`g'(t)=[(t^2+3)(2t)-t^2(2t)]/(t^2+3)^2=0`

` ` `2t^3+6t-2t^3=0`

`6t=0`

`t=0`

Plug in the critical t value(s) and the endpoints of the closed interval into the g(t) function.

`g(t)=t^2/(t^2+3)`

`g(-1)=1/4`

`g(0)=0`

`g(1)=1/4`

Examine the g(t)...

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Given: `g(t)=(t^2)/(t^2+3)[-1,1]`

Find the critical values for t by setting the derivative equal to zero and solving for the t value(s).

`g'(t)=[(t^2+3)(2t)-t^2(2t)]/(t^2+3)^2=0`

` ` `2t^3+6t-2t^3=0`

`6t=0`

`t=0`

Plug in the critical t value(s) and the endpoints of the closed interval into the g(t) function.

`g(t)=t^2/(t^2+3)`

`g(-1)=1/4`

`g(0)=0`

`g(1)=1/4`

Examine the g(t) values to determine the absolute extrema.

The absolute maximum is the points are  `(-1, 1/4), (1,1/4).`

The absolute minimum is the point (0, 0).

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