Find the absolute maximum value and the absolute minimum value, if any, of the function. (If an answer does not exist, enter DNE.)
f(x) = −x2 + 2x + 3 on [3, 6]
maximum
minimum
find f(3) and f(6)
find where f'(x) = 0 for local extrma in the interval
pick largest/smallest of those values
To find the absolute maximum and minimum values of a function on a given interval, we can follow these steps:
1. Find all critical points of the function within the given interval. Critical points occur where the derivative is either zero or undefined.
2. Evaluate the function at these critical points as well as the endpoints of the interval.
3. The highest value among these evaluated points will be the absolute maximum (if it exists), and the lowest value will be the absolute minimum (if it exists).
Let's follow these steps to find the maximum and minimum values for the function f(x) = -x^2 + 2x + 3 on the interval [3, 6]:
Step 1: Find the critical points within the interval [3, 6]:
To find the critical points, we need to find where the derivative of the function is zero or undefined.
The derivative of f(x) with respect to x is given by f'(x) = -2x + 2.
To find where f'(x) = 0, we set -2x + 2 = 0 and solve for x:
-2x + 2 = 0
-2x = -2
x = 1
So, we have a critical point at x = 1.
Step 2: Evaluate the function at the critical point and endpoints:
Evaluate f(x) at the critical point x = 1:
f(1) = -(1)^2 + 2(1) + 3
f(1) = -1 + 2 + 3
f(1) = 4
Evaluate f(x) at the endpoints of [3, 6]:
f(3) = -(3)^2 + 2(3) + 3
f(3) = -9 + 6 + 3
f(3) = 0
f(6) = -(6)^2 + 2(6) + 3
f(6) = -36 + 12 + 3
f(6) = -21
Step 3: Determine the absolute maximum and minimum values:
From the evaluations above, we have the following values:
f(1) = 4
f(3) = 0
f(6) = -21
The highest value among these is 4, which occurs at x = 1. So, the absolute maximum value is 4.
The lowest value among these is -21, which occurs at x = 6. So, the absolute minimum value is -21.
Therefore, the absolute maximum value is 4, and the absolute minimum value is -21 for the function f(x) = -x^2 + 2x + 3 on the interval [3, 6].