Find the maximum value of g(t)=t^3−9t+7 on the interval [−1.5,3].

Question

Find the maximum value of g(t)=t^3-9t+7 on the interval [-1.5,3] .

Answer 1

Use a first derivative test to determine whether there are any critical points for g on the interval.

1 Find the derviative.

g'(t) = 3t² - 9

2 Find roots of the derivative.

t = ±√3

3 Determine whether these critical points are (1) in the interval and (2) maxima, minima, or neither using the first derivative test sign chart.

-√3 is not in the interval, but √3 is.

<____(-)______0_____(+)_____>

0 √3 2

This is a minimum because the function is decreasing (- rate of change) before the critical point and increasing (+ rate of change) after. When this happens, you'll just test the endpoints of the interval and see which one produces a maximum value for the function (not the derivative).

4. Find g(-1.5) and g(3).

g(3) < g(-1.5) and g(-1.5) = 17.125, the maximum value of g on this interval.