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Question 997829: Let f(x) = 2x^2 - 9x Find all extreme values (if any) of f on the interval [0, 9]
So far I have:
f'(x) = 4x - 9
f(x) = x = 4/9
If I plug back into the original function I can find either the max or min value.
f(4/9) = 2(4/9)^2 - 9(4/9) = - 10.125 (calculator)
However, how do I know whether this is the max or min value without a graphical refrence?
Also, isn't there a rule with extrema stating that since the interval notation is [0,9] that I can use the 9? if so why not the 0. This rule keeps tripping me up.
Please explain
Thank you
Found 2 solutions by Alan3354, solver91311:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! So far I have:
f'(x) = 4x - 9
---
At the min or max, f'(x) = 0
4x-9 = 0
x = 9/4
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f(9/4) = 2*81/16 - 9*9/4
= 81/8
===========================
It's a parabola in x. The x^2 term is positive --> opens upward
--> a minimum.
The 1st derivative is linear --> 1 extreme.
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f(x) = x = 4/9
If I plug back into the original function I can find either the max or min value.
f(4/9) = 2(4/9)^2 - 9(4/9) = - 10.125 (calculator)
However, how do I know whether this is the max or min value without a graphical refrence?
Also, isn't there a rule with extrema stating that since the interval notation is [0,9] that I can use the 9? if so why not the 0. This rule keeps tripping me up.
------------
9/8 is between 0 and 9, so that's not an issue.
Answer by solver91311(24713)  (Show Source):
You can put this solution on YOUR website!
If you have a local extremum on the interval, then there will be a point in the interval where the first derivative is equal to zero, which is to say:
Hint: It is NOT 4/9.
If the second derivative is positive at that point, then the extremum is a minimum, if the second derivative is negative, the extremum is a maximum.
As for the endpoints of the interval, you need to find the value of the function at each of the endpoints and compare those function values to the function values at your other critical points. Basically, you have three critical points, one of which is a minimum, one of which is a maximum, and the other is somewhere in between.
John
My calculator said it, I believe it, that settles it
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