Reverse a function on a suitable domain

Try

A machine computes f(x)=3x+2. You observe output 17 and want the original input. Undo the addition, then undo the multiplication: (17−2)/3=5. Test by sending 5 through the original machine.

Now replace the machine by f(x)=x². Observing output 25 no longer identifies a unique real input: both 5 and −5 work. An inverse relation can list both possibilities, but an inverse function must assign a single recovered input to each permitted output.

Predict

Predict what information would make the squaring machine reversible. “The input is nonnegative” works; “the output is nonnegative” adds nothing because every squared real output already is. A restriction must remove the ambiguity among original inputs.

Experiment

Consider f(x)=(x−1)² on x≥1. Build a table and reverse its columns.

Original input x	Original output y	Recovered input 1+√y
1	0	1
2	1	2
4	9	4
6	25	6

Plot the original points and reversed points on equal-scale axes. Reversing coordinates reflects the graph across y=x. The original domain [1,∞) becomes the inverse range, while the original range [0,∞) becomes the inverse domain.

Repeat the construction with x≤1 instead. The inverse is now 1−√y. Both restrictions are legitimate, but they select different original inputs for the same positive output.

Understand

A function is one-to-one, or injective, when different allowed inputs always give different outputs. A horizontal line meeting its graph at two distinct points reveals a collision. This test concerns the graph’s chosen domain, not every possible input of the formula.

To find an inverse algebraically, write y=f(x), solve for x in terms of y, then rename the inverse input if desired. For y=(2x−5)/3, multiplying by 3 and adding 5 gives x=(3y+5)/2. Thus f⁻¹(u)=(3u+5)/2 on the reals. The superscript −1 in f⁻¹ denotes an inverse function, not a reciprocal; 1/f(x) is a different operation.

Verification needs both compositions on their correct domains. f⁻¹(f(x)) must return each original input x. Also f(f⁻¹(y)) must return each allowed inverse input y. For unrestricted squaring and the proposed inverse √y, the second composition works for y≥0, but the first gives √(x²)=|x|. It fails to recover negative x. One successful composition does not establish an unrestricted inverse.

Even an injective rule may not cover a declared codomain. The inverse is naturally defined on the original range. For example, f(x)=2x with domain [0,1] reaches only [0,2], even if its declared codomain is all real numbers. An inverse input of 5 is not permitted for that restricted function.

Apply

For every inverse question, record original domain, original range, inverse domain and inverse range. Verify a symbolic formula instead of relying only on a table. In the square-error response, name the collision and show which composition fails.

The logarithm investigation offers another inverse pair with explicit positive-base conditions. The principle of recovering an input is shared, even though exponential and quadratic restrictions differ.

Continue

Systems ask for inputs satisfying multiple constraints simultaneously. Their graph intersections can be numerous, unique or absent, so the next lesson retains all candidate pairs and checks them in every original equation.