Question

Find the maximum value of g(t)=x^3−9x+4 on the interval [−1.5,3]

Answer 1

g(x) = x^3-9x+4

Answer 2

take the derivative, set it equal to zero and solve
that will give you the critical points
then test \(g(whatever)\), g(-1.5)) and \(g(3)\)
the biggest is the max and the smallest is the min

Answer 3

you can almost find the critical point in your head
\[g'(x)=3x^2-9=3(x^2-3)=3(x+\sqrt3)(x-\sqrt3)\]