Let f(x)=-x^2+3x on the interval [1,3]. Find the absolute maximum and absolute minimum of f(x) on this interval.
take f ' (x)
set it equal to zero and solve for x
(I see x = 3/2)
plug that back into f(x) and evaluate
Also plug in the end values of your domain
pick the smallest and largest values
I got my maximum as 3/2 but cannot find the minimum that is in the closed interval (1,3)
3/2 is not the maximum f(3/2) is the local maximum
f(3/2) = -(9/4) + 3(3/2) = 9/4 or 2.25
checking the endvalues of your domain
f(1) = -1+3 = 2
f(3) = -9 + 9 = 0
so within your domain, the absolute min is 0 and the absolute max is 2.25
look at the graph from Wolfram
http://www.wolframalpha.com/input/?i=plot+-x%5E2%2B3x+%2C+1%3Cx%3C3
To find the absolute maximum and absolute minimum of a function on a closed interval, we need to follow these steps:
Step 1: Identify the critical points of the function on the given interval. These are the points where the derivative is either zero or undefined.
To find the critical points, we need to compute the derivative of the function f(x) with respect to x.
f(x) = -x^2 + 3x
f'(x) = -2x + 3
To find the critical points, we set f'(x) equal to zero and solve for x:
-2x + 3 = 0
2x = 3
x = 3/2
So, x = 3/2 is the only critical point in the interval [1,3].
Step 2: Check the function values at the critical points and the endpoints of the interval.
The critical point x = 3/2 is within the interval [1,3].
Now, we evaluate f(x) at the critical point and endpoints:
f(1) = -(1)^2 + 3(1) = -1 + 3 = 2
f(3/2) = -(3/2)^2 + 3(3/2) = -9/4 + 9/2 = -9/4 + 18/4 = 9/4 (or 2.25)
f(3) = -(3)^2 + 3(3) = -9 + 9 = 0
Step 3: Compare the function values to determine the absolute maximum and minimum.
From the calculations, we have the following function values:
f(1) = 2
f(3/2) = 9/4 (or 2.25)
f(3) = 0
The absolute maximum is the highest function value, which is f(3/2) = 9/4 (or 2.25).
The absolute minimum is the lowest function value, which is f(3) = 0.
Therefore, the absolute maximum is 9/4 (or 2.25) and the absolute minimum is 0 on the interval [1,3].