Melody

\(\text{Find the invers of }f(x)=x*|x|\)

If x>=0  then  f(x)=x^2  which is a very simple  half concave up parabola in the first quadrant

if X < 0  then  f(x)=-x^2  which is a very simple  half concave down parabola in the third quadrant

To be honest, the first thing I did was sketch this. The inverse is the refection in the line y=x so I can already see where this is going.

But lets look at the algebra

let   y= f(x)  )it is just easier to work with)

If x >=0     then y is also positive

y=x^2

x=+sqrt(y)

so

f^-1(x)=+sqrt(x)

If  x<0   then y is also negative

x=-sqrt(x)

so

f^-1(x)=-sqrt(|x|)

The tricky bit was how to put these together into one function   

|x| / x = 1    if x>0        and    |x| / x = -1    if x<0     so

\(f^{-1}(x)=\frac{|x|*\sqrt{|x|}}{x}\)  

check:

You need to work out a logical system

id say  1+1+1+1+5     that is the only one with 4 ones

then     1+1+1+(two that sum to 6 but neither can be 1)

then    1+1+(three that sum to 7 but not including an more ones)

etc

when you exhaust all the possible combinations that inclu....

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thinking about it is is preobably easier to go from the big end.

The biggest number you can have is 5  (there is only the one possible way)

Now look at all possibilities that include fours.

You just need to find a logical system to list them all so that none will get forgotten or duplicated.

No not all.

Sorry.   I answered this yesterday but it had been autoblocked and  I did not realize.

Actually, looking again, I do not know.

Certainly all of them can be true whether or not they need to be true is a different matter.

I am withdrawing my earlier assertion that one can be easily removed as the answer. I misread it.

Thinking about it now one at a time.

I think all of them have to be true.   

Do you mean D is the only one that is right or the only one that can easily be dismissed?

Your shapes are right.

The most general one is the Trapezoid  (In Australia we would call it a trapezium)

But unfortunately, your conclusion is not correct.   (unless I have made some really stupid error)

Draw your trapezoid and re-check your result     

Draw a pic. of a random quadrilateral.   One that does not look very special

Call it PQRS

Mark on it the things that are told to you.

Redraw it so that it looks more like it is supposed to.  But do not give it extra special features that are no specified (unless it is unavoidable)

What shape is it?

Now test all those possibilities and you will see that only one seems right.

This is a good question.

One choice can be dismissed easily enough, which would that be, and why?

\(f(x) = \frac{x - \sqrt{3}}{x\sqrt{3} + 1} \\~\\ f(f(x)) = \frac{\frac{x - \sqrt{3}}{x\sqrt{3} + 1} - \sqrt{3}}{\frac{x - \sqrt{3}}{x\sqrt{3} + 1}\sqrt{3} + 1}\\ \text{This simplifies to}\\ f(f(x)) =\frac{x+\sqrt3}{1-\sqrt3x}\\~\\ f(f(f(x)) )=\frac{\frac{x+\sqrt3}{1-\sqrt3x}+\sqrt3}{1-\sqrt3\frac{x+\sqrt3}{1-\sqrt3x}}\\~\\ \text{This simplifies to}\\ f(f(f(x)) )=x\\~\\ \text{now there is a loop happening.}\\~\\ f^0(x)=x\\ f^1(x) = \frac{x - \sqrt{3}}{x\sqrt{3} + 1} \\~\\ f^2(x)=\frac{x+\sqrt3}{1-\sqrt3x}\\~\\ f^3(x)=x\\~\\ \)

2012=2(mod3)

so it  seems that

\(f^{2012}(x)=\frac{x+\sqrt3}{1-\sqrt3x}\\~\\\)

Coding

f(x) = \frac{x - \sqrt{3}}{x\sqrt{3} + 1} \\~\\

f(f(x)) = \frac{\frac{x - \sqrt{3}}{x\sqrt{3} + 1} - \sqrt{3}}{\frac{x - \sqrt{3}}{x\sqrt{3} + 1}\sqrt{3} + 1}\\
\text{This simplifies to}\\
f(f(x)) =\frac{x+\sqrt3}{1-\sqrt3x}\\~\\

f(f(f(x)) )=\frac{\frac{x+\sqrt3}{1-\sqrt3x}+\sqrt3}{1-\sqrt3\frac{x+\sqrt3}{1-\sqrt3x}}\\~\\
\text{This simplifies to}\\
f(f(f(x)) )=x\\~\\
\text{now there is a loop happening.}\\~\\
f^0(x)=x\\
f^1(x) = \frac{x - \sqrt{3}}{x\sqrt{3} + 1} \\~\\

f^2(x)=\frac{x+\sqrt3}{1-\sqrt3x}\\~\\

f^3(x)=x\\~\\