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 + 3 x^2 - 1 on [-3, 1]
f(-3) = -1
f(1) = 3
Now we have the max/min at the endpoints. There may be other, different min/max inside the interval.
f'(x) = 3x^2 + 6x = 3x(x+2)
so,
f(-2) = 3 is a local max
f(0) = -1 is a local min
On the interval [-3,1] the absolute
max is 3
min is -1
To find the absolute maximum and minimum values of the function f(x) = x^3 + 3x^2 - 1 on the interval [-3, 1], we need to find the critical points and endpoints.
1. First, let's find the critical points by taking the derivative of f(x) and setting it equal to zero:
f'(x) = 3x^2 + 6x = 0
Simplifying, we get:
3x(x + 2) = 0
Setting each factor equal to zero, we find two critical points:
x = 0 and x = -2.
2. Next, we evaluate the function at these critical points and the endpoints of the interval [-3, 1]:
f(0) = (0)^3 + 3(0)^2 - 1 = -1
f(-2) = (-2)^3 + 3(-2)^2 - 1 = -9
3. Now, we need to check the function at the endpoints:
f(-3) = (-3)^3 + 3(-3)^2 - 1 = -25
f(1) = (1)^3 + 3(1)^2 - 1 = 3
4. Finally, we compare the function values at the critical points and endpoints to determine the absolute maximum and minimum values:
The absolute maximum value is 3, which occurs at x = 1.
The absolute minimum value is -25, which occurs at x = -3.
To find the absolute maximum and absolute minimum values of a function on a closed interval, we need to evaluate the function at its critical points and endpoints.
Step 1: Find the critical points of the function.
To find the critical points, we need to find where the derivative of the function is zero or undefined. Let's find the derivative of f(x):
f(x) = x^3 + 3x^2 - 1
Taking the derivative:
f'(x) = 3x^2 + 6x
Setting f'(x) = 0 to find the critical points:
3x^2 + 6x = 0
3x(x + 2) = 0
Setting each factor equal to zero:
3x = 0 or x + 2 = 0
Solving for x:
x = 0 or x = -2
So, the critical points of the function are x = 0 and x = -2.
Step 2: Evaluate the function at the critical points and endpoints.
Now, we need to evaluate the function at the critical points and endpoints of the interval [-3, 1].
Let's find f(-3):
f(-3) = (-3)^3 + 3(-3)^2 - 1
f(-3) = -27 + 27 - 1
f(-3) = -1
Now, let's find f(1):
f(1) = (1)^3 + 3(1)^2 - 1
f(1) = 1 + 3 - 1
f(1) = 3
Finally, let's find f(0):
f(0) = (0)^3 + 3(0)^2 - 1
f(0) = -1
Step 3: Compare the values to find the absolute maximum and minimum.
Now that we have evaluated the function at the critical points and endpoints, we can compare these values to find the absolute maximum and minimum.
f(x) = x^3 + 3x^2 - 1
f(-3) = -1
f(0) = -1
f(1) = 3
So, the absolute maximum value of the function is 3, which occurs at x = 1, and the absolute minimum value is -1, which occurs at x = -3 and x = 0.
Therefore, the absolute maximum value is 3, and the absolute minimum value is -1.