Suppose a function f(x)=2x3−3x2−36x+12 is defined on the interval [−5,5] . What is the global minimum of the function f(x) ? (1 point) Responses −133 − 133 −56 − 56 56 56 −69 − 69 69
To find the global minimum of the function, we need to find the critical points of the function, and then evaluate the function at those points.
To find the critical points, we take the derivative of the function: f'(x) = 6x^2 - 6x - 36.
To find when f'(x) = 0, we set the derivative equal to zero and solve for x:
6x^2 - 6x - 36 = 0
Factoring the equation, we get:
6(x^2 - x - 6) = 0
Setting each factor equal to zero, we get two possible solutions:
x^2 - x - 6 = 0
(x - 3)(x + 2) = 0
x = 3 or x = -2
Now we evaluate the function at the critical points and endpoints of the interval.
f(-5) = 2(-5)^3 - 3(-5)^2 - 36(-5) + 12 = -133
f(3) = 2(3)^3 - 3(3)^2 - 36(3) + 12 = -69
f(5) = 2(5)^3 - 3(5)^2 - 36(5) + 12 = 56
f(-2) = 2(-2)^3 - 3(-2)^2 - 36(-2) + 12 = 56
Therefore, the global minimum of the function f(x) is -133.
What is the global maximum of the function f(x) ? (1 point) Responses 72 72 −56 − 56 56 56 −69 − 69 69
To find the global maximum of the function, we can again evaluate the function at the critical points and endpoints of the interval.
f(-5) = 2(-5)^3 - 3(-5)^2 - 36(-5) + 12 = -133
f(3) = 2(3)^3 - 3(3)^2 - 36(3) + 12 = -69
f(5) = 2(5)^3 - 3(5)^2 - 36(5) + 12 = 56
f(-2) = 2(-2)^3 - 3(-2)^2 - 36(-2) + 12 = 56
Therefore, the global maximum of the function f(x) is 56.