SOLUTION: Consider the function f(x) = x-3ln(x), 1/3 <= x <= 9
The absolute maximum value is ___ at x = ___
The absolute minimum value is ___ at x = ___
I know I need to find the CP
Algebra.Com
Question 1204526: Consider the function f(x) = x-3ln(x), 1/3 <= x <= 9
The absolute maximum value is ___ at x = ___
The absolute minimum value is ___ at x = ___
I know I need to find the CP
f'(x) = 1 - 3/x
1-3/x = 0
1 = 1/3x
x = 3
and there are the end points at 0 and 9
f(0) = 1/3 - 3ln(1/3) = 1/3 + 3ln(3)
f(3) = 3 - 3ln(3)
f(9) = 9 - 3ln(9) = 9 - 6ln(3)
However, I put these answers in and the autograder says they are incorrect and any help is appreciated :)
Found 3 solutions by ikleyn, MathLover1, greenestamps:
Answer by ikleyn(52832) (Show Source): You can put this solution on YOUR website!
.
In your solution (in your post), you say that the end points are at 0 and 9.
It is . The endpoints are 1/3 and 9.
Your work should be re-done.
Answer by MathLover1(20850) (Show Source): You can put this solution on YOUR website!
your interval is NOT
GLOBAL (ABSOLUTE) MINIMA
(,)=(,)
GLOBAL (ABSOLUTE) MAXIMA
(,)=(,)
Answer by greenestamps(13203) (Show Source): You can put this solution on YOUR website!
In the work you show, you show endpoints of 0 and 9, when in fact they are 1/3 and 9. Where you say your are evaluating f(0), you are in fact correctly evaluating f(1/3).
And you don't show us what answers you actually put in that the autograder said were not correct.
There is only one absolute minimum and only one absolute maximum; we can't tell from your post what you said they were.
The minimum is at x=3; and f(1/3) is greater than f(9), so the maximum is at x=1/3.
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