Let f(x)=-x^2+3x on the interval [1,3]. Find the absolute maximum and absolute minimum of f(x) on this interval.
The absolute max occurs at x=.
The absolute min occurs at x=
at 1, y = 2
at 3, y = 0
dy/dx = -2x + 3 = 0 if x = 3/2
at x = 3/2, y = 2.25
so max at x = 3/2
and
min at x = 3
To find the absolute maximum and absolute minimum of the function f(x)=-x^2+3x on the interval [1,3], we can follow these steps:
1. Calculate the values of f(x) at the critical points within the interval [1,3]. These critical points occur when the derivative of f(x) is zero or undefined.
To find the critical points, we need to find the derivative of f(x). The derivative of f(x) with respect to x is given by f'(x) = -2x + 3.
Setting f'(x) = 0, we have -2x + 3 = 0.
Adding 2x to both sides, we get 3 = 2x.
Dividing both sides by 2, we find x = 3/2. Therefore, the critical point within the interval [1,3] is x = 3/2.
2. Evaluate f(x) at the endpoints of the interval [1,3], which are x = 1 and x = 3.
Now, let's plug in the values of x into the function f(x) to calculate the corresponding y-values.
When x = 1, f(1) = -(1^2) + 3(1) = 2.
When x = 3, f(3) = -(3^2) + 3(3) = 0.
So we have two potential candidates:
Candidate 1: (1, 2)
Candidate 2: (3, 0)
3. Compare the values of f(x) at the critical points and endpoints to determine the absolute maximum and absolute minimum.
Comparing the values, we see that (1, 2) represents the absolute maximum because f(1) = 2 is greater than both f(3) = 0 and f(3/2).
Similarly, (3, 0) represents the absolute minimum because f(3) = 0 is less than both f(1) = 2 and f(3/2).
Therefore, the absolute maximum occurs at x = 1, and the absolute minimum occurs at x = 3.