T.6 · Functions, inverse branches and composition

A function assigns exactly one output to each allowed input. Injective means equal outputs force equal inputs; surjective means every element of the stated codomain is reached. The map x↦x² from R to [0,∞) is surjective but not injective. From [0,∞) to [0,∞) it is both, hence bijective. From [0,∞) to R it remains injective but fails surjectivity. Keep domain, image and codomain separate when testing each claim.

An inverse function reverses a bijection. Solve y=f(x) for x, then use the original domain to choose a branch. For f(x)=(x−2)²−5 on x≥2, the inverse is 2+√(y+5) on y≥−5. The minus branch would return inputs outside the chosen domain. Check f⁻¹(f(x))=x for allowed x and f(f⁻¹(y))=y for allowed y; one unchecked composition can conceal a domain error.

The inverse graph reflects the original graph across y=x; it does not take reciprocals of output values. A self-inverse function, or involution, satisfies f(f(x))=x wherever the composition is defined. Both −x on R and 1/x on R excluding zero are involutions. A strictly increasing involution on an interval must be the identity: if f(x)>x, increasingness gives f(f(x))>f(x)>x, and the analogous argument rules out f(x)<x.

Iteration fⁿ means repeated composition, not the power (f(x))ⁿ. Calculate the first few compositions and check their domains before looking for a period. For f(x)=1/(1−x) on R excluding 0 and 1, f²(x)=(x−1)/x and f³(x)=x. The image stays in the same allowed domain. Therefore reduce an iteration count modulo three. A displayed formula equal to x after cancellation does not restore forbidden inputs.

For f(x)=(x−2)²−5 with x≥2, f(5)=4 and f⁻¹(4)=2+√9=5. The unrestricted quadratic would have two inputs, −1 and 5, giving output 4. For g(x)=1/(1−x), the orbit 3→−1/2→2/3→3 has period three, so g⁸(3)=2/3; the count concerns compositions, not eighth powers.

Changing a codomain changes surjectivity. Reflection across y=x is an inverse graph; reciprocating y-values is a different operation. Do not cancel away excluded inputs.

Restrict $f(x)=(x-1)^2+2$ to $x\ge1$. State its image and inverse, including domains. Verify both inverse compositions.

Worked solution

The image is $[2,\infty)$. Solving $y=(x-1)^2+2$ with $x\ge1$ selects $x=1+\sqrt{y-2}$. Thus $f^{-1}:[2,\infty)\to[1,\infty)$ has that formula. For allowed x, $1+\sqrt{(x-1)^2}=x$ because $x-1\ge0$. For allowed y, $(\sqrt{y-2})^2+2=y$. The negative square-root branch belongs to a different restriction, not a second inverse on this domain.

Let $h(x)=1/(1-x)$. Determine the domain of $h\circ h\circ h$ and its formula. May the simplified expression be extended to the excluded points without changing the original composite?

Check after attempting

The first stage requires $x\ne1$. The second requires $h(x)\ne1$, which excludes x=0. For $x\ne0,1$, $h^2(x)=(x-1)/x$. This never equals one, so the third stage adds no exclusion. Then $h^3(x)=x$ on $\mathbb R\setminus\{0,1\}$. The identity formula is defined more widely, but adding 0 or 1 defines an extension, not the original composite. Each intermediate input must be permitted before cancellation.