Porisma

Worked solution

Undo a sequence in reverse order

A graph y=f(x)y=f(x) undergoes three successive transformations: A, a scaling parallel to the x-axis with scale factor 1/21/2; B, a translation of 3 units in the positive x-direction; C, a reflection in the x-axis. The resulting equation is y=−ln⁡(6−2x)y=-\ln(6-2x). Determine the equation of the graph y=f(x)y=f(x), showing your working clearly.

[2006/CJC/Promo/3(modified)]

CJC · Transformations · Example 12

Method 1: undo C, B, then A

Answer

y=ln⁡(−x)y=\ln(-x).

Reverse both the order and each operation. The Teacher asks that the replacements be shown when necessary. Start from the resulting curve, not from the unknown original:

Undo the reflection

−y=−ln⁡(6−2x)⟹y=ln⁡(6−2x)-y=-\ln(6-2x)\quad\Longrightarrow\quad y=\ln(6-2x)

C is its own inverse: replace y by −y-y.

Undo the right translation

y=ln⁡(6−2(x+3))=ln⁡(−2x)y=\ln\bigl(6-2(x+3)\bigr)=\ln(-2x)

Translate left three: replace x by x+3x+3.

Undo the horizontal scaling

y=ln⁡ ⁣(−2(x2))=ln⁡(−x)y=\ln\!\left(-2\left(\frac x2\right)\right)=\ln(-x)

Scale parallel to the x-axis by factor 2: replace x by x/2x/2.

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Method 2: follow an arbitrary point

Answer

q=ln⁡(−p)q=\ln(-p), hence y=ln⁡(−x)y=\ln(-x).

The distinct Teacher alternative follows a point (p,q)(p,q) on the original graph. After A it is (p/2,q)(p/2,q); after B, (p/2+3,q)(p/2+3,q); after C, (p/2+3,−q)(p/2+3,-q). Substitute into the known final equation: −q=−ln⁡(6−2(p/2+3))-q=-\ln\bigl(6-2(p/2+3)\bigr), so q=ln⁡(−p)q=\ln(-p). Independently, the real domain is x<0x<0 for the original, mapping to x<3x<3 for the final curve; this is a domain check, not an extra printed ask.

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