Find the absolute maximum and absolute minimum values of f on the given interval.
f(x) = (x2 - 1)3
[-1, 6]
To find the absolute maximum and absolute minimum values of a function on a given interval, we can follow these steps:
Step 1: Find the critical points by finding where the derivative of the function is equal to zero or does not exist.
Step 2: Evaluate the function at the critical points and the endpoints of the interval.
Step 3: The largest function value is the absolute maximum, and the smallest function value is the absolute minimum.
Let's follow these steps for the given function f(x) = (x^2 - 1)^3 on the interval [-1, 6].
Step 1: Find the critical points
To do this, we need to find where the derivative of the function is equal to zero or does not exist. Differentiating f(x), we get:
f'(x) = 3(x^2 - 1)^2 * 2x = 6x(x^2 - 1)^2
To find the critical points, we set f'(x) = 0 and solve for x:
6x(x^2 - 1)^2 = 0
Either 6x = 0 or (x^2 - 1)^2 = 0
Solving 6x = 0, we find x = 0.
To solve (x^2 - 1)^2 = 0, we take the square root of both sides:
x^2 - 1 = 0
x^2 = 1
x = ±1
So, the critical points are x = -1, x = 0, and x = 1.
Step 2: Evaluate the function at the critical points and endpoints
Now, we evaluate the function at the critical points and the endpoints of the given interval [-1, 6].
f(-1) = ((-1)^2 - 1)^3 = (1 - 1)^3 = 0^3 = 0
f(6) = ((6)^2 - 1)^3 = (36 - 1)^3 = 35^3 = 42,875
Step 3: Determine the absolute maximum and absolute minimum values
Comparing the function values, we see that:
- The absolute maximum value is 42,875, which occurs at x = 6.
- The absolute minimum value is 0, which occurs at x = -1.
Therefore, the absolute maximum and absolute minimum values of f(x) = (x^2 - 1)^3 on the interval [-1, 6] are 42,875 and 0, respectively.
To find the absolute maximum and absolute minimum values of f(x) = (x^2 - 1)^3 on the interval [-1, 6], we need to first find the critical points and the endpoints of the interval.
1. Find the critical points:
To find the critical points, we need to set the derivative of f(x) equal to zero and solve for x.
f'(x) = 3(x^2 - 1)^2 * 2x
= 6x(x^2 - 1)(x^2 - 1)
Setting f'(x) equal to zero:
6x(x^2 - 1)(x^2 - 1) = 0
From this equation, we can see that f'(x) = 0 when x = 0 or x^2 - 1 = 0.
Solving x^2 - 1 = 0:
x^2 - 1 = 0
(x - 1)(x + 1) = 0
x - 1 = 0 or x + 1 = 0
x = 1 or x = -1
Therefore, the critical points on the interval [-1, 6] are x = -1 and x = 1.
2. Evaluate f(x) at the critical points and endpoints:
Now we need to evaluate f(x) at x = -1, x = 1, and the endpoints -1 and 6.
f(-1) = (-1^2 - 1)^3 = (1 - 1)^3 = 0^3 = 0
f(1) = (1^2 - 1)^3 = (1 - 1)^3 = 0^3 = 0
f(-1) = (6^2 - 1)^3 = (36 - 1)^3 = 35^3 = 42,875
3. Compare the values obtained:
The values obtained are:
f(-1) = 0
f(1) = 0
f(6) = 42,875
Therefore, the absolute maximum value of f(x) on the interval [-1, 6] is 42,875, and the absolute minimum value is 0.