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

This Question is Closed

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0then lets get the derivative :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[9\ln x6/x^2\] i think this is it

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0(x)(9/x)  (3+9ln(x))(1)  x^2 3(2 3ln(x))  x^2

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0I need help? like professional help?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0.....me and my voices are content :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok so now we need to find the absolute max by setting f' to 0

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.00 will give us critical points, that we can test for minand max

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Do you have a particular bound in which you're looking for the max, or is it over (infinity, +infinity)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0polpak is also good at this stuff :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well.. 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
 5 years ago
Best ResponseYou've already chosen the best response.0if I see it right: 2 = 3ln(x) is what we are looking to solve

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes that was correct...lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0cbrt(e^2) is a possible critical point.... the only one I can come up with

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0as x gets large, this goes to zero I think

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the 2nd derivative I believe is: 3(7+6ln(x))  x^3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x=e^(3/2) would be the max value. As it approaches 0 as x goes to zero or infinity.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i finished that problem i got a different one.... mean slope=10 f'x=6x^212x90 mean value theroem

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0at 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
 5 years ago
Best ResponseYou've already chosen the best response.0finished? kaputs? done? howd we do :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the answer was x=e^(2/3)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0hmmm.... I was right :) I like being right. even when im wrong...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0whats that new problem? cant quite decipher it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok it was like another we did the average or mean slope =10 and my f'x=6x^212x90 I need to find values that make this equal to the slope

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0let find the derivative that equals that slope to get the equation for the tangent line from it :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That is the derivative...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0ohh.... 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
 5 years ago
Best ResponseYou've already chosen the best response.06x^212x90 = 10 6x^2 12x 80 = 0 6(x^2 2x  40/3) nice numbers :) remember to stop me when I do something stupid :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0do quad formula on it to speed things up :) 144(4)(80)(6) 144 + 1920 12 + sqrt(2064)  right? 12

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0since the original will be a cubic, it stands to reason that there are 2 places that give us a slope of 10

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0in fact, we can just suit up f'(x) to get that function right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0(S) 6x^212x90 dx 2x^3 6x^2 90x +c is in the family of curves

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0we can drop the constant if we want to just use the x values to get it... right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[2(3x^2 6x 40) = 0\] \[\rightarrow 3x 6x 40 = 0\] \[\rightarrow x = \frac{6 \pm \sqrt{516}}{6}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So at those x values the slope is 10.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0should be.... if I didnt mess it up :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well I did it myself, so if your x matches mine it's probably right.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0I see the 1 :) can 516 get reduced?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0close.... I mighta messed something up tho :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The 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: \[f'(x)=\frac{69 \text{Log}(x)}{x^2} \] Set the Numerator to zero and solve for x \[69 \text{Log}(x)=0\] \[x\text{=}e^{2/3} \] 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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