Locate the absolute extrema of the function f(x)=2x^2-12x+4 on the closed interval [-6,6].
To locate the absolute extrema of a function on a closed interval, you typically need to follow these steps:
1. Find the critical points of the function within the interval.
2. Evaluate the function at the critical points and at the endpoints of the interval.
3. Identify the highest and lowest function values to determine the absolute maximum and minimum.
Let's follow these steps for the given function, f(x) = 2x^2 - 12x + 4, on the interval [-6, 6]:
1. Find the critical points:
To find the critical points, we need to find where the derivative of the function is equal to zero or does not exist.
Taking the derivative of f(x), we get: f'(x) = 4x - 12.
Now, set f'(x) = 0 and solve for x:
4x - 12 = 0
4x = 12
x = 3
So, we have a critical point at x = 3.
2. Evaluate the function at the critical point and the endpoints:
Now, we need to evaluate the function at the critical point and at the endpoints of the interval [-6, 6].
At x = -6: f(-6) = 2(-6)^2 - 12(-6) + 4 = 72 + 72 + 4 = 148
At x = 6: f(6) = 2(6)^2 - 12(6) + 4 = 72 - 72 + 4 = 4
At x = 3 (critical point): f(3) = 2(3)^2 - 12(3) + 4 = 18 - 36 + 4 = -14
3. Identify the highest and lowest function values:
From the evaluations, we see that:
- The function value at x = -6 is 148.
- The function value at x = 6 is 4.
- The function value at x = 3 is -14.
Therefore, the maximum value of the function on the closed interval [-6, 6] is 148, and it occurs at x = -6.
The minimum value of the function on the closed interval is -14, and it occurs at x = 3.
So, the location of the absolute extrema on the closed interval [-6, 6] for the function f(x) = 2x^2 - 12x + 4 is as follows:
Absolute maximum: (x = -6, f(x) = 148)
Absolute minimum: (x = 3, f(x) = -14)