Business Calculus Question

Question

Consider the function 𝑓(𝑥)=5−3𝑥^2, −3 ≤ x ≤ 2

The absolute maximum value is:

and this occurs at 𝑥x equals:

The absolute minimum value is:

and this occurs at 𝑥x equals:

Answer 1

f(x) = 5-3x^2

f'(x) = -6x = 0, x=0 is a turning point, a max or min, the vertex of a downward opening parabola, the maximum point, local and global

f(0) =5 -3(0)^2 = 5, the point (0,5)

f(-3) =5-3(-3)^2 = 5-27 =-22 the absolute minimum in the interval -3<x<2 when x=-3, the point (-3, -22)

f(2) = 5-3(2)^2 = 5-12 = -7