Find the absolute extrema of the function on the closed interval.
f (x) = x^3 − 12x, [0, 4]
minimum (x, y) =( )
maximum (x, y) = ( )
I can't even give a setup cause I am lost please help!
f(x) = x^3 - 12x. You can find the maxima and minima by differentiating it (I'm assuming here that you can use calculus) to get f'(x) = 3x^2 - 12, and solving f'(x) = 0, so 3x^2 - 12 = 0, so 3x^2 = 12, so x^2 = 4, so x is either +2 or -2. But only one of those lies in the interval [0, 4], namely x = +2. The original function has at most two turning points (because that's the order of the derivative function), and only one of them lies within the interval, so any other maxima or minina must lie at the ends of the interval, i.e. either x=0 or x=4. At x=0, f(x) = 0. At x=4, f(x) = 64 - 48 = 16. At x=2, f(x) = 8 - 24 = -16. So the minimum must be -16 at x=2, and the maximum is +16 at x=4. Done.
Thank you :)
To find the absolute extrema of a function on a closed interval, we need to identify the critical points and endpoints of the interval and determine the values of the function at these points. The absolute extrema will be the maximum and minimum values among these points.
1. To start, let's find the critical points by finding the derivative of the function.
f'(x) = 3x^2 - 12
2. Setting the derivative equal to zero, we can solve for x.
3x^2 - 12 = 0
3(x^2 - 4) = 0
(x^2 - 4) = 0
(x - 2)(x + 2) = 0
Thus, the critical points are x = 2 and x = -2.
3. Now we need to evaluate the function at the endpoints and the critical points to find the absolute extrema.
- At x = 0 (the left endpoint of the interval),
f(0) = (0)^3 - 12(0) = 0.
- At x = 4 (the right endpoint of the interval),
f(4) = (4)^3 - 12(4) = 64 - 48 = 16.
- At x = 2 (the critical point),
f(2) = (2)^3 - 12(2) = 8 - 24 = -16.
- At x = -2 (the critical point),
f(-2) = (-2)^3 - 12(-2) = -8 + 24 = 16.
4. Comparing the values found, we can identify the minimum and maximum values.
The minimum value occurs at the point (2, -16).
The maximum value occurs at the points (4, 16) and (-2, 16).
So, the answer is:
minimum (x, y) = (2, -16)
maximum (x, y) = (4, 16) and (-2, 16)
To find the absolute extrema of a function on a closed interval, you need to follow the steps below:
1. Find all critical points of the function within the interval by taking the derivative of the function and set it equal to zero. Critical points could also occur at the endpoints of the interval.
2. Evaluate the function at each critical point and the endpoints to find the corresponding function values.
3. Compare all the function values obtained in step 2 to determine the absolute minimum and maximum.
Let's apply these steps to your specific function f(x) = x^3 − 12x on the interval [0, 4].
1. Find the derivative of f(x):
f'(x) = 3x^2 - 12
Now, set f'(x) = 0 and solve for x to find the critical points:
3x^2 - 12 = 0
3(x^2 - 4) = 0
(x - 2)(x + 2) = 0
=> x = 2 or x = -2
2. Evaluate f(x) at each critical point and the endpoints:
- Endpoint 1: x = 0
f(0) = (0)^3 - 12(0) = 0
- Endpoint 2: x = 4
f(4) = (4)^3 - 12(4) = 64 - 48 = 16
- Critical point 1: x = 2
f(2) = (2)^3 - 12(2) = 8 - 24 = -16
- Critical point 2: x = -2
f(-2) = (-2)^3 - 12(-2) = -8 + 24 = 16
3. Compare the function values obtained:
- Minimum (x, y): (2, -16)
- Maximum (x, y): (4, 16)
Therefore, the absolute minimum occurs at (2, -16), and the absolute maximum occurs at (4, 16) for the given function on the closed interval [0, 4].