Proving Inequality for Linear Functions: |h(h(x))+h(h(1/x))|>2

[utkarshakash]

Homework Statement

If f and g are two distinct linear functions defined on R such that they map[-1,1] onto [0,2] and h:R-{-1,0,1}→R defined by h(x)=f(x)/g(x) then show that |h(h(x))+h(h(1/x))|>2

Homework Equations

The Attempt at a Solution

I assume f(x) to be ax+b and g(x) to be lx+m so that h(x) is (ax+b)/(lx+m). From here I can write h(h(x)) and h(h(1/x)) but there is nothing I can see that will help me to prove this inequality. Any ideas?

 

[SammyS]

Homework Statement

If f and g are two distinct linear functions defined on R such that they map[-1,1] onto [0,2] and h:R-{-1,0,1}→R defined by h(x)=f(x)/g(x) then show that |h(h(x))+h(h(1/x))|>2

Homework Equations

The Attempt at a Solution

I assume f(x) to be ax+b and g(x) to be lx+m so that h(x) is (ax+b)/(lx+m). From here I can write h(h(x)) and h(h(1/x)) but there is nothing I can see that will help me to prove this inequality. Any ideas?

You can be more specific regarding the functions, f & g .

There are only two distinct linear functions which map [-1,1] onto [0,2] .

What are they?

 

[utkarshakash]

You can be more specific regarding the functions, f & g .

There are only two distinct linear functions which map [-1,1] onto [0,2] .

What are they?

y=x+1 and y=-x+1. Are these correct?

 

[SammyS]

y=x+1 and y=-x+1. Are these correct?

Yes. Equivalently, y=1+x and y=1-x .

So, there are only two cases to consider.

Choosing f(x) = 1+x and g(x) = 1-x, for now, what is h(1/x) ?

 

[utkarshakash]

Yes. Equivalently, y=1+x and y=1-x .

So, there are only two cases to consider.

Choosing f(x) = 1+x and g(x) = 1-x, for now, what is h(1/x) ?

x+1/x-1 which can be reduced to -f(x)/g(x)

And after simplifying further I am left with proving this inequality

|x-(1/x)|>2

 

[SammyS]

x+1/x-1 which can be reduced to -f(x)/g(x)

It looks like you get h(1/x) = -f(x)/g(x), for either way of assigning 1-x and 1+x to f(x) and g(x).

And after simplifying further I am left with proving this inequality

|x-(1/x)|>2

I got something similar to x-(1/x), but it is different.

Check your algebra.

It seems to me that one could do this more abstractly using some calculus, etc. (Maybe the Calculus part comes in the next step)