Question

if f(x)=1-x/1+x,x>0 then least value of f(f(x)) + f(f(1/x)) is what

Answer 1

need some paper for this one

Answer 2

is if its own inverse? looks like it

Answer 3

so
\[f(f(x))=x\]

Answer 4

and likewise
\[f(f(\frac{1}{x})=\frac{1}{x}\]

Answer 5

so you are trying to minimize
\[x+\frac{1}{x},x>0\]

Answer 6

minimum is 2 if x = 1

Answer 7

you need the method for this or is it clear?

Answer 8

i need the method

Answer 9

ok first is it clear that
\[f(f(x))=x\]?

Answer 10

yes

Answer 11

the fuction is its own inverse??

Answer 12

then it should also be clear that
\[f(f(\frac{1}{x}))=\frac{1}{x}\]

Answer 13

well yes, if
\[f(f(x))=x\] that means the function is its own inverses

Answer 14

yes

Answer 15

understood thnx

Answer 16

ok so our real job once the algebra is out of the way, is to find the minimum value of
\[x+\frac{1}{x}\] for x > 0
there are two ways to do this, one is calculus and one is thinking. is this a calc course?

Answer 17

We can use AM>=GM

Answer 18

i guess so. if you use calc you take the derivative, find that it is
\[1-\frac{1}{x^2}\] find the critical points where the derivative is 0 i.e. solve
\[1-\frac{1}{x^2}=0\]
\[\frac{1}{x^2}=1\]
\[x=\pm1\]and since we know x> 0 we know x = 1, and we also know it is a minimum, and
\[1+\frac{1}{1}=2\]

Answer 19

think method is that this is symmetric in that if you replace x by 1/x you get the same thing back again

Answer 20

meaning the minimum value has to occur when x and 1/x are equal, meaning they are both 1, and so you get 2

Answer 21

and now that you mentioned it, using AM >= GM you see that
\[\frac{1}{2}(x+\frac{1}{x})\geq \sqrt{x\times \frac{1}{x}}\]
\[\frac{1}{2}(x+\frac{1}{x})\geq \sqrt{1}=1\] and so
\[x+\frac{1}{x}\geq 2\]

Answer 22

and since we can make it 2, by making x = 1 we know that this is a minimum
i have to say i would not have thought of that one!!