Find the absolute maximum value and the absolute minimum value, if any, of the function. (If an answer does not exist, enter DNE.)
f(x)=x^3+3x^2-1 on [-3,1]
f(-3) = -1
f(1) = 3
f'(x) = 3x^2 + 6x = 3x(x+2)
so, there are relative max/min at x= -2,0
f(-2) = 3
f(0) = -1
So, on [-3,1] we have
min = -1
max = 3
To find the absolute maximum and minimum values of the function f(x) = x^3 + 3x^2 - 1 on the interval [-3, 1], we can follow these steps:
1. Calculate the critical points by finding the derivative of f(x) and setting it equal to zero:
f'(x) = 3x^2 + 6x = 0
To solve for x, we factor out the common factor, which is x:
x(3x + 6) = 0
Setting each factor equal to zero gives us two possible critical points:
x = 0 and x = -2
2. Evaluate the function f(x) at the critical points and the endpoints of the interval:
- Plug in x = -3, -2, 0, and 1 into f(x).
f(-3) = (-3)^3 + 3(-3)^2 - 1 = -19
f(-2) = (-2)^3 + 3(-2)^2 - 1 = -1
f(0) = (0)^3 + 3(0)^2 - 1 = -1
f(1) = (1)^3 + 3(1)^2 - 1 = 3
3. Compare the values obtained. The largest value represents the absolute maximum, and the smallest value represents the absolute minimum.
Based on the calculations:
f(-3) = -19
f(-2) = -1
f(0) = -1
f(1) = 3
The absolute maximum value is 3, and it occurs at x = 1.
The absolute minimum value is -19, and it occurs at x = -3.