anonymous
  • anonymous
f(x)=(3+9ln(x))/x i need to find the x coordinate of the absolute max...
Mathematics
katieb
  • katieb
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amistre64
  • amistre64
then lets get the derivative :)
anonymous
  • anonymous
\[-9\ln x-6/x^2\] i think this is it
anonymous
  • anonymous
u need help

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amistre64
  • amistre64
(x)(9/x) - (3+9ln(x))(1) ------------------- x^2 3(2 -3ln(x)) ----------- x^2
anonymous
  • anonymous
help me
amistre64
  • amistre64
I need help? like professional help?
anonymous
  • anonymous
lol
amistre64
  • amistre64
.....me and my voices are content :)
anonymous
  • anonymous
ok so now we need to find the absolute max by setting f' to 0
amistre64
  • amistre64
0 will give us critical points, that we can test for minand max
anonymous
  • anonymous
Do you have a particular bound in which you're looking for the max, or is it over (-infinity, +infinity)
amistre64
  • amistre64
polpak is also good at this stuff :)
anonymous
  • anonymous
x>0
anonymous
  • anonymous
Well.. What is the limit as x approaches infinity is another thing to consider then when looking for absolute max. You have to check for critical points, but also end points. If the function goes to infinity as x goes to infinity then it has no absolute max value.
amistre64
  • amistre64
if I see it right: 2 = 3ln(x) is what we are looking to solve
anonymous
  • anonymous
yes that was correct...lol
amistre64
  • amistre64
cbrt(e^2) is a possible critical point.... the only one I can come up with
amistre64
  • amistre64
as x gets large, this goes to zero I think
anonymous
  • anonymous
Yes, it does.
amistre64
  • amistre64
the 2nd derivative I believe is: 3(7+6ln(x)) --------- x^3
anonymous
  • anonymous
x=e^(3/2) would be the max value. As it approaches 0 as x goes to zero or infinity.
anonymous
  • anonymous
i finished that problem i got a different one.... mean slope=-10 f'x=6x^2-12x-90 mean value theroem
amistre64
  • amistre64
at cbrt(e^2) the 2nd derivative would be 33/e^2; which is positive.... I believe that it is a MIN at that point
amistre64
  • amistre64
finished? kaputs? done? howd we do :)
anonymous
  • anonymous
the answer was x=e^(2/3)
amistre64
  • amistre64
hmmm.... I was right :) I like being right. even when im wrong...
amistre64
  • amistre64
whats that new problem? cant quite decipher it
anonymous
  • anonymous
Ok it was like another we did the average or mean slope =-10 and my f'x=6x^2-12x-90 I need to find values that make this equal to the slope
amistre64
  • amistre64
let find the derivative that equals that slope to get the equation for the tangent line from it :)
anonymous
  • anonymous
That is the derivative...
amistre64
  • amistre64
ohh.... my eyes are going :) make it equal to -10; then add 10 to both sides to get it equal to 0 again and sove for x right?
amistre64
  • amistre64
6x^2-12x-90 = -10 6x^2 -12x -80 = 0 6(x^2 -2x - 40/3) nice numbers :) remember to stop me when I do something stupid :)
amistre64
  • amistre64
do quad formula on it to speed things up :) 144-(4)(-80)(6) 144 + 1920 12 +- sqrt(2064) --------------- right? 12
amistre64
  • amistre64
1 +- sqrt(129)/3
amistre64
  • amistre64
since the original will be a cubic, it stands to reason that there are 2 places that give us a slope of -10
amistre64
  • amistre64
in fact, we can just suit up f'(x) to get that function right?
amistre64
  • amistre64
(S) 6x^2-12x-90 dx 2x^3 -6x^2 -90x +c is in the family of curves
amistre64
  • amistre64
we can drop the constant if we want to just use the x values to get it... right?
anonymous
  • anonymous
\[2(3x^2 -6x -40) = 0\] \[\rightarrow 3x -6x -40 = 0\] \[\rightarrow x = \frac{6 \pm \sqrt{516}}{6}\]
anonymous
  • anonymous
So at those x values the slope is -10.
amistre64
  • amistre64
should be.... if I didnt mess it up :)
anonymous
  • anonymous
Well I did it myself, so if your x matches mine it's probably right.
amistre64
  • amistre64
I see the 1 :) can 516 get reduced?
amistre64
  • amistre64
maybe to 129?
amistre64
  • amistre64
close.... I mighta messed something up tho :)
anonymous
  • anonymous
The equation is: \[f(x)=\frac{3+9 \text{Log}(x)}{x}\] The first derivative, \[f'(x)=\frac{9}{x^2}-\frac{3+9 \text{Log}(x)}{x^2} \] simplified, is: \[f'(x)=\frac{6-9 \text{Log}(x)}{x^2} \] Set the Numerator to zero and solve for x \[6-9 \text{Log}(x)=0\] \[x\text{=}e^{2/3} \] Pluggin x into f(x) gives the coordinates of the maximum funtion value: \[\left\{e^{2/3},\frac{9}{e^{2/3}}\right\}\text{ = }\{1.94773,4.62075\} \] Refer to the attachment plot of f(x).

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