Question

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

Answer 1

then lets get the derivative :)

Answer 2

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

Answer 3

u need help

Answer 4

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

3(2 -3ln(x))
-----------
x^2

Answer 5

help me

Answer 6

I need help? like professional help?

Answer 7

lol

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

polpak is also good at this stuff :)

Answer 13

x>0

Answer 14

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.

Answer 15

if I see it right:

2 = 3ln(x) is what we are looking to solve

Answer 16

yes that was correct...lol

Answer 17

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

Answer 18

as x gets large, this goes to zero I think

Answer 19

Yes, it does.

Answer 20

the 2nd derivative I believe is:

3(7+6ln(x))
---------
x^3

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

finished? kaputs? done? howd we do :)

Answer 25

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

Answer 26

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

Answer 27

whats that new problem? cant quite decipher it

Answer 28

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

Answer 29

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

Answer 30

That is the derivative...

Answer 31

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?

Answer 32

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 :)

Answer 33

do quad formula on it to speed things up :)

144-(4)(-80)(6)

144 + 1920

12 +- sqrt(2064)
--------------- right?
12

Answer 34

1 +- sqrt(129)/3

Answer 35

since the original will be a cubic, it stands to reason that there are 2 places that give us a slope of -10

Answer 36

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

Answer 37

(S) 6x^2-12x-90 dx

2x^3 -6x^2 -90x +c is in the family of curves

Answer 38

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

Answer 39

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

Answer 40

So at those x values the slope is -10.

Answer 41

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

Answer 42

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

Answer 43

I see the 1 :) can 516 get reduced?

Answer 44

maybe to 129?

Answer 45

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

Answer 46

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).