Find the absolute maximum and absolute minimum values of f on the given interval.
f(x) = 17 + 4x − x^2, [0, 5]
I got -13 and 17, but the answers were wrong. Take the derivative for the critical values, right? To get 2 as a critical value. What from there?
f' = 0 = 4 - 2 x
so x = 2, right
at 2
y = 17 + 8 - 4 = 21
check end points
at x = 0, y = 17
at x = 5, y = 17+20 -25 =12
21 is at the vertex and is maximum
12 is minimum on this restricted domain
To find the absolute maximum and minimum values of the function f(x) on the interval [0, 5], we can follow these steps:
1. Calculate the derivative of f(x) with respect to x.
f'(x) = 4 - 2x
2. Find the critical values by setting f'(x) equal to zero and solving for x.
We have: 4 - 2x = 0
Solving this equation gives x = 2 as a critical value.
3. Identify the endpoints of the given interval [0, 5]. These are x = 0 and x = 5.
4. Evaluate the function f(x) at the critical values and endpoints.
- Evaluate f(x) at x = 0:
f(0) = 17 + 4(0) - (0)^2
f(0) = 17
- Evaluate f(x) at x = 2:
f(2) = 17 + 4(2) - (2)^2
f(2) = 17 + 8 - 4
f(2) = 21
- Evaluate f(x) at x = 5:
f(5) = 17 + 4(5) - (5)^2
f(5) = 17 + 20 - 25
f(5) = 12
5. Compare the values obtained from step 4:
The absolute maximum value of f(x) is the largest value among the evaluated values, which is 21 at x = 2.
The absolute minimum value of f(x) is the smallest value among the evaluated values, which is 12 at x = 5.
Therefore, the correct absolute maximum and minimum values of f(x) on the interval [0, 5] are 21 and 12 respectively.