A self-inverse map and a restricted modulus inverse

The source graph has asymptotes $x=3$, $y=3$, and axial intercepts $(7/3,0)$ and $(0,7/3)$. Each horizontal line meets it at most once: rewriting $f(x)=3+2/(x-3)$ shows that the left branch has outputs below 3, the right branch outputs above 3, and each branch decreases strictly. Thus $f$ is one-to-one. Let $y=f(x)$, so $f^{-1}(y)=x$. Then $(3-x)y=7-3x$, $3y-xy=7-3x$, $x(3-y)=7-3y$, and $x=(7-3y)/(3-y)$. Rename the inverse input to obtain the stated rule. Its domain is $R_f=(-\infty,3)\cup(3,\infty)=D_f$. Rule and domain agree, proving self-inverse behaviour.

Since $f=f^{-1}$, $ff=ff^{-1}$ gives $f^2(x)=x$. Then $ff^2(x)=f(x)$ gives $f^3(x)=f(x)$, and $ff^3(x)=ff(x)$ gives $f^4(x)=x$. Continuing, $f^{2002}(x)=x$, so $f^{2003}(x)=ff^{2002}(x)=f(x)$. Hence $f^{2003}(5)=f(5)=(7-3(5))/(3-5)=4$. The superscript counts compositions; it is not a pointwise power. Independently 5 and 4 form a two-cycle and neither hits the excluded input 3.

Choose the equation before rearranging, as the Teacher note stresses. On $x\le-1$, $(2-x)(1+x)\le0$, so $y=-(2-x)(1+x)=-(2+x-x^2)=x^2-x-2$. The unrestricted modulus illustration has roots −1,2 and central maximum $(1/2,9/4)$; only its left branch belongs to $g$. Set $y=g(x)$, so $g^{-1}(y)=x$. Complete the square: $y=[(x-1/2)^2-1/4]-2$, hence $(x-1/2)^2=y+9/4$, $x-1/2=\pm\sqrt{y+9/4}$, and $x=1/2\pm\sqrt{y+9/4}$. Reject the positive branch because the original inputs satisfy $x\le-1$. Thus $g^{-1}(y)=1/2-\sqrt{y+9/4}$; rename its input to $x$. The original left branch has range $[0,\infty)$, so that is the inverse domain, not merely the larger algebraic radical domain $[-9/4,\infty)$.

Sketch $g(x)=x^2-x-2$ only for $x\le-1$, with closed endpoint $(-1,0)$ and no y-intercept. Reflect every point across $y=x$ to obtain $g^{-1}$ on $x\ge0$, starting at closed $(0,-1)$ and decreasing. It has no x-intercept because all outputs are at most −1. Draw $y=x$ and use equal axis scales to make the reflection explicit. A vertical axis marked x in the source sketch is a label slip; the reconstruction uses y.

$gg^{-1}(x)=x$ on the entire original domain of the inner inverse: $D_{gg^{-1}}=D_{g^{-1}}=[0,\infty)$. The inverse outputs are accepted by $g$. Do not extend the identity beyond this domain.

$R_{g^{-1}}=(-\infty,-1]\subseteq D_f=(-\infty,3)\cup(3,\infty)$. Every inverse output avoids the excluded input 3, so $fg^{-1}$ exists on all of $D_{g^{-1}}=[0,\infty)$.

Use the source mapping method: $D_{g^{-1}}=[0,\infty)$ maps through $g^{-1}$ to $(-\infty,-1]$, then through $f$ to $[5/2,3)$. The question asks for the range, so no composite rule is needed. At the included endpoint, $f(-1)=(7-3(-1))/(3-(-1))=10/4=5/2$. As the intermediate input tends to $-\infty$, $f(t)=3+2/(t-3)$ approaches 3 from below without reaching it. On $t\le-1$ the function decreases as t increases; it reaches every output between those bounds. Include 5/2 and exclude 3.