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 ->  Customizable Word Problem Solvers  -> Misc -> 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       Log On

Ad: Over 600 Algebra Word Problems at edhelper.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(52831) About Me  (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  highlight%28highlight%28FATALLY%29%29  highlight%28highlight%28INCORRECT%29%29.   The endpoints are  1/3  and  9.


Your work should be re-done.



Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

your interval is 1%2F3+%3C=+x+%3C=+9+NOT 0+%3C=+x+%3C=+9+

GLOBAL (ABSOLUTE) MINIMA
(x,f%28x%29)=(3,3-3ln%283%29)
GLOBAL (ABSOLUTE) MAXIMA
(x,f%28x%29)=(1%2F3,1%2F3%2B3ln%283%29)
download

Answer by greenestamps(13203) About Me  (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.