# `h(t) = t/(t + 3), [-1,6]` Find the absolute extrema of the function on the closed interval. Given: `h(t)=t/(t+3),[-1,6]`

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

`h'(t)=[(t+3)(1)-t(1)]/(t+3)^2=0`

`t+3-t=0`

`3=0`

A critical value will not exists for this function.

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

`h(t)=t/(t+3)`

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Given: `h(t)=t/(t+3),[-1,6]`

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

`h'(t)=[(t+3)(1)-t(1)]/(t+3)^2=0`

`t+3-t=0`

`3=0`

A critical value will not exists for this function.

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

`h(t)=t/(t+3)`

`h(-1)=-1/2`

`h(6)=2/3`

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

The absolute maximum is the point `(6,2/3).`

The absolute minimum is the point `(-1,-1/2)`

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