# Joselito’s Clay Molding Technique

Engr. Joselito Torculas. Pioneered JMCT

If AoPS has SFFT, we also have Joselito’s Clay Molding Technique ( JCMT ). Before anyone will name this technique, I decided to publish this and name it after the person who popularized the method.

JCMT is a technique first used by Engr. Joselito Torculas, one of the admins of Elite Math Circle. The technique is used in composition of functions. He described the method as “shaping the clay to its form”. Thus the name is JCMT.

Illustrative Examples:

Worked Example 1:

Let $g(x)=2x-1$  and  $(f\circ g)(x)=3x^2-1$. What is $f(x)$?

Solution: By JCMT,

We know that $(f\circ g)(x)=f(g(x))$

But     $(f\circ g)(x)=3x^2-1$,

$3x^2-1=f(g(x))$

$f(g(x))= 3x^2-1$

But     $g(x)=2x-1$,

$f(2x-1)= 3x^2-1$

Now, since we are just looking for $f(x)$, we need to look for a value of x to make 2x-1 become x. This can be easily accomplished using the inverse of 2x-1 but we will not use that.

The first stage of JCMT is to eliminate the constant in 2x-1 and eventually x and to the desired function.

Let    $x=x+\displaystyle\frac{1}{2}$

$f(2(x+0.5)-1)=3(x+\displaystyle\frac{1}{2})^2-1$

$f(2x+1-1)=3(x+\displaystyle\frac{1}{2})^2-1$

$f(2x)= 3(x+\displaystyle\frac{1}{2})^2-1$

Let    $x=\displaystyle\frac{x}{2}$

$f(2(\displaystyle\frac{x}{2}))=3((\displaystyle\frac{x}{2})+\displaystyle\frac{1}{2})^2-1$

$f(x)=3(\displaystyle\frac{x^2+2x+1}{4})-1$

$f(x)=\displaystyle\frac{3x^2+6x-1}{4}$

Worked Example 2:

Given that $f(2x-4)=\displaystyle\frac{2x-1}{x+4}$. What is the value of $f(x-1)?$

Solution: By JCMT,

Let    $x=x+2$

$f(2x-4)=\displaystyle\frac{2x-1}{x+4}$

$f(2(x+2)-4)=\displaystyle\frac{2(x+2)-1}{x+2+4}$

$f(2x+4-4)=\displaystyle\frac{2x+4-1}{x+2+4}$

$f(2x)=\displaystyle\frac{2x+3}{x+6}$

Let    $x=\displaystyle\frac{x}{2}$

$f(2x)=\displaystyle\frac{2x+3}{x+6}$

$f(2(\displaystyle\frac{x}{2}))=\displaystyle\frac{2(\displaystyle\frac{x}{2})+3}{(\displaystyle\frac{x}{2})+6}$

$f(x)=\displaystyle\frac{x+3}{\frac{x+12}{2}}$

$f(x)=\displaystyle\frac{2(x+3)}{x+12}$

Finally, let    $x=x-1$

$f(x-1)=\displaystyle\frac{2(x-1+3)}{x-1+12}$

$f(x-1)=\displaystyle\frac{2(x+2)}{x+11}$

Worked Example 3:

If $f(2x+1)=3x^2-1$. What is $f(x^2)?$

Solution: By JCMT

Let    $x=x-\displaystyle\frac{1}{2}$

$f(2x+1)=3x^2-1$

$f(2(x-\displaystyle\frac{1}{2})+1)=3(x-\displaystyle\frac{1}{2})^2-1$

$f(2x+1-1)=3(x-\displaystyle\frac{1}{2})^2-1$

$f(2x)=3(x-\displaystyle\frac{1}{2})^2-1$

Let    $x=\displaystyle\frac{x^2}{2}$

$f(2\cdot \displaystyle\frac{x^2}{2})=3(\displaystyle\frac{x^2}{2}-\displaystyle\frac{1}{2})^2-1$

$f(x^2)=3(\displaystyle\frac{x^2-1}{2})^2-1$

$f(x^2)=3(\displaystyle\frac{x^4-2x^2+1}{4})-1$

$f(x^2)=3(\displaystyle\frac{x^4-2x^2+1}{4})-1$

$f(x^2)=\displaystyle\frac{3x^4-6x^2+3-4}{4}$

$f(x^2)=\displaystyle\frac{3x^4-6x^2-1}{4}$

### 3 Responses

1. John Cyril says:

for example in #2, why not equate directly x=(x+4)/2 to get f(x), same as to #3, equate directly x=(x-1)/2 to get f(x)? Like for #3, if you will equate it x=(x-1)/2 directly to the RHS, f(x)=3[(x-1)/2]^2 -1 = (3x^2 -6x -1)/4 so, f(x)=(3x^4 -6x^2 -1)/4 and it will yield the same answer as yours…

2. Right! But not all people can figure that out especially to those who new to this. 🙂 One step at a time. Thanks John! 🙂

3. ayos_dito says:

Minor comment: On Example 1, the dot sign (.) was incorrectly used to denote composition of functions. Dot usually means multiplication. The symbol used for composition is usually a circle (circ in latex, IIRC).

Otherwise, very good method. I usually just use the inverse of the function inside the composition, but I can see how this could be easier and more elegant in some cases. Keep it up!