# `h(s) = 1/(s - 2), [0,1]` Find the absolute extrema of the function on the closed interval.

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Given: `h(s)=1/(s-2),[0,1]`

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

`h'(s)=[(s-2)0-1(1)]/(x-2)^2=0`

`1=0`

A critical value will not exist for this function.

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

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

` h(0)=-1/2 `

`h(1)=-1`

Examine the h(s) values to determine the **absolute extrema**.

The **absolute maximum** is the point **(0, -1/2)**.

The **absolute minimum** is the point **(1, -1)**.

You need to find the derivative of the function, using the quaotient rule, such that:

`h'(s) = (1'*(s - 2) - 1*(s -2)')/((s-2)^2)`

`h'(s) = (0*(s - 2) - 1*1)/((s-2)^2)`

`h'(s) =1/((s-2)^2)`

You need to solve for s the equation h'(s) = 0, such that:

`1/((s-2)^2) = ` 0

Notice that there exists no values of s to satisfy the equation.

**Hence, evaluating the absolute extrema of the given function, there exists no such values.**