Find the local and absolute extreme values of the function on the given interval. f(x)=x^3-6x^2+9x+1, [2,4]?
To find the local and absolute extreme values of the function f(x) = x^3 - 6x^2 + 9x + 1 on the given interval [2, 4], we can follow these steps:
1. Determine the critical points:
To find the critical points, we need to find the values of x where the derivative of the function is equal to zero or does not exist.
2. Find the first derivative:
f'(x) = 3x^2 - 12x + 9
3. Set the first derivative equal to zero and solve for x:
3x^2 - 12x + 9 = 0
Factoring the equation, we have:
(x - 1)(3x - 9) = 0
Solving for x, we get:
x - 1 = 0 => x = 1
3x - 9 = 0 => x = 3
So, the critical points are x = 1 and x = 3.
4. Find the second derivative:
f''(x) = 6x - 12
5. Determine the nature of the critical points:
Plug the critical points into the second derivative to determine the nature of the critical points.
For x = 1:
f''(1) = 6 * 1 - 12 = -6
Since f''(1) is negative, it means x = 1 is a local maximum point.
For x = 3:
f''(3) = 6 * 3 - 12 = 6
Since f''(3) is positive, it means x = 3 is a local minimum point.
6. Evaluate the function at the critical points and the endpoints of the interval:
f(2) = 2^3 - 6(2)^2 + 9(2) + 1 = 1
f(4) = 4^3 - 6(4)^2 + 9(4) + 1 = 17
7. Compare the function values at the critical points and endpoints:
The function values are as follows:
f(2) = 1
f(4) = 17
f(1) = 5
f(3) = -1
- The local maximum value is f(1) = 5.
- The local minimum value is f(3) = -1.
- The absolute maximum value is f(4) = 17.
- The absolute minimum value is f(3) = -1.
Therefore, the local and absolute extreme values of the function f(x) = x^3 - 6x^2 + 9x + 1 on the interval [2, 4] are:
- Local maximum value: 5 at x = 1.
- Local minimum value: -1 at x = 3.
- Absolute maximum value: 17 at x = 4.
- Absolute minimum value: -1 at both x = 3 and x = 2 (endpoints of the interval).
To find the local and absolute extreme values of a function on a given interval, you need to follow these steps:
Step 1: Find the critical points.
To find the critical points, you need to calculate the derivative of the function and solve for x when the derivative is equal to zero or undefined.
Given function:
f(x) = x^3 - 6x^2 + 9x + 1
Taking the derivative of f(x):
f'(x) = 3x^2 – 12x + 9
Setting the derivative equal to zero to find critical points:
3x^2 – 12x + 9 = 0
Apply the quadratic formula to solve for x:
x = (-(-12) ± sqrt((-12)^2 - 4(3)(9))) / (2(3))
x = (12 ± sqrt(144 - 108)) / 6
x = (12 ± sqrt(36)) / 6
x = (12 ± 6) / 6
x = (18 / 6) or (6 / 6)
x = 3 or 1
So, the critical points are x = 3 and x = 1.
Step 2: Evaluate the function at the critical points and the endpoints of the given interval.
Evaluate f(x) at x = 2, x = 3, and x = 4 to find the local and absolute extreme values.
When x = 2:
f(2) = (2)^3 - 6(2)^2 + 9(2) + 1
f(2) = 8 - 24 + 18 + 1
f(2) = 3
When x = 3:
f(3) = (3)^3 - 6(3)^2 + 9(3) + 1
f(3) = 27 - 54 + 27 + 1
f(3) = 1
When x = 4:
f(4) = (4)^3 - 6(4)^2 + 9(4) + 1
f(4) = 64 - 96 + 36 + 1
f(4) = 5
Step 3: Determine the local and absolute extreme values.
Compare the values of f(x) at the critical points and endpoints.
The function f(x) has the following values:
f(2) = 3
f(3) = 1
f(4) = 5
Comparing these values, we can see that the local maximum value is 5 at x = 4, and the local minimum value is 1 at x = 3.
Since we are only considering the interval [2, 4], the absolute maximum value is 5 at x = 4 (the local maximum) and the absolute minimum value is 1 at x = 3 (the local minimum).
f(x)=x^3-6x^2+9x+1, [2,4]
f' = 3 x^2 -12 x + 9
look for zeros in domain
x^2 -4x + 3 = 0
(x-3)(x-1) = 0
x = 1 or x = 3
3 is in our piece of domain between 2 and 4
What is function at x = 3 ? Calculate it.
Then also calculate the function at x = 2 and x = 4