f(x)=(3+9ln(x))/x
i need to find the x coordinate of the absolute max...

- anonymous

f(x)=(3+9ln(x))/x
i need to find the x coordinate of the absolute max...

- katieb

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- amistre64

then lets get the derivative :)

- anonymous

\[-9\ln x-6/x^2\] i think this is it

- anonymous

u need help

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## More answers

- amistre64

(x)(9/x) - (3+9ln(x))(1)
-------------------
x^2
3(2 -3ln(x))
-----------
x^2

- anonymous

help me

- amistre64

I need help? like professional help?

- anonymous

lol

- amistre64

.....me and my voices are content :)

- anonymous

ok so now we need to find the absolute max by setting f' to 0

- amistre64

0 will give us critical points, that we can test for minand max

- anonymous

Do you have a particular bound in which you're looking for the max, or is it over (-infinity, +infinity)

- amistre64

polpak is also good at this stuff :)

- anonymous

x>0

- 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

if I see it right:
2 = 3ln(x) is what we are looking to solve

- anonymous

yes that was correct...lol

- amistre64

cbrt(e^2) is a possible critical point.... the only one I can come up with

- amistre64

as x gets large, this goes to zero I think

- anonymous

Yes, it does.

- amistre64

the 2nd derivative I believe is:
3(7+6ln(x))
---------
x^3

- anonymous

x=e^(3/2) would be the max value. As it approaches 0 as x goes to zero or infinity.

- anonymous

i finished that problem i got a different one....
mean slope=-10 f'x=6x^2-12x-90
mean value theroem

- 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

finished? kaputs? done? howd we do :)

- anonymous

the answer was x=e^(2/3)

- amistre64

hmmm.... I was right :) I like being right. even when im wrong...

- amistre64

whats that new problem? cant quite decipher it

- 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

let find the derivative that equals that slope to get the equation for the tangent line from it :)

- anonymous

That is the derivative...

- 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

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

do quad formula on it to speed things up :)
144-(4)(-80)(6)
144 + 1920
12 +- sqrt(2064)
--------------- right?
12

- amistre64

1 +- sqrt(129)/3

- 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

in fact, we can just suit up f'(x) to get that function right?

- amistre64

(S) 6x^2-12x-90 dx
2x^3 -6x^2 -90x +c is in the family of curves

- amistre64

we can drop the constant if we want to just use the x values to get it... right?

- anonymous

\[2(3x^2 -6x -40) = 0\]
\[\rightarrow 3x -6x -40 = 0\]
\[\rightarrow x = \frac{6 \pm \sqrt{516}}{6}\]

- anonymous

So at those x values the slope is -10.

- amistre64

should be.... if I didnt mess it up :)

- anonymous

Well I did it myself, so if your x matches mine it's probably right.

- amistre64

I see the 1 :) can 516 get reduced?

- amistre64

maybe to 129?

- amistre64

close.... I mighta messed something up tho :)

- 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).

##### 1 Attachment

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