`h(s) = 1/(s - 2), [0,1]` Find the absolute extrema of the function on the closed interval.
Find the critical values for s by setting the derivative equal to zero and solving for the s value(s).
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(0)=-1/2 `
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)`
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.