Question

$$ \text { If } \large f(x)= \frac { (x+ \frac 1x)^6 + (x^6+ \frac 1{x^6})-12}{(x+ \frac 1x)^4 + (x^3+ \frac 1{x^3})} $$ $$\text{ and x > 0, find the maximum value of f(x) }$$

Answer 1

wow

Answer 2

Calculus free please!

Answer 3

Else it would be too tedious..

Answer 4

lol I won't even dare to do it with Calculus

Answer 5

Me 2 :D

Answer 6

just plug it in wolfram, and let it compute the maximum!

Answer 7

3 is the answer

Answer 8

My calculator is crunching for the solution.

Answer 9

lol

Answer 10

Wow , you guys are really lethargic :P

Answer 11

How am I even supposed to start this problem, Hit and Trial would be useless

Answer 12

yes it will be I suppose some sort of trying to reduce as perfect squares then cancelling might help ..

Answer 13

try using substitution

Answer 14

Hey I think I got it please don't post the solution

Answer 15

I think the whole problem is a red herring

Answer 16

there is no global maximum... but how do we show that?

Answer 17

this one?

Answer 18

there are minima tho

Answer 19

at x = 1, the expression is 3....

Answer 20

now shower me with medalz!!!

Answer 21

is 3 the right answer

Answer 22

3 is the right answer.... for the global minimum when x > 0... and not for the maximum (which does not really exist)

Answer 23

this resembles x^6/x^4: x^2 to me .... but thats only at large values of x

which means there is no max since x^2 goes to infinity right?

Answer 24

right @amistre

Answer 25

but why won't you let us use calculus to figure that out :(

Answer 26

and what defines calculus?

Answer 27

check this out,
at x =1 denominator attains minima

so it is a possible answer

Answer 28

calculus is the field of professional mathematics which involves coming up with new integration rules....

Answer 29

@stom

and the value of the function at x=1 is 3 :-D

Answer 30

http://www.wolframalpha.com/input/?i=%28%28%28x%2B1%2Fx%29^6%2B%28x^6%2B1%2F%28x^6%29%29-12%29%2F%28%28x%2B1%2Fx%29^4%2B%28x^3%2B%281%2F%28x^3%29%29%29%29

Answer 31

i know its 3 at x=1,
thats why i said 3 is the answer,

Answer 32

at the very bottom it shows the global minimum, and a little before that it becomes quite apparent that there is no maximum.

Answer 33

i didn't get it :( is hit and trial the only way

Answer 34

there is no maxima ,

Answer 35

take t = x+1/x

Answer 36

\[\frac{\left(x + \frac{1}{x}\right)^6 + \left(x + \frac{1}{x}\right)\left[\left( x + \frac{1}{x}\right)^2 -3\right] - 14}{\left( x + \frac{1}{x}\right)\left[\left(x + \frac{1}{x}\right)^3 + \left(x + \frac{1}{x}\right)^2 - 3\right]}\]


This is as far as I can get :-/

Answer 37

Whoopsy

\[\frac{\left(x + \frac{1}{x}\right)^6 + \left(x + \frac{1}{x}\right)^2\left[\left( x + \frac{1}{x}\right)^2 -3\right]^2 - 14}{\left( x + \frac{1}{x}\right)\left[\left(x + \frac{1}{x}\right)^3 + \left(x + \frac{1}{x}\right)^2 - 3\right]}\]

Answer 38

so what is the correct way to solve this @ffm?

Answer 39

Yeah I am curious too