Invert a restricted logarithm and carry its range into a cubic

Method 1: the graph of the restricted logarithm passes every horizontal-line test, so $f$ is one-to-one. Method 2: $f\prime(x)=1/(x-a)>0$ throughout $(a,a+2]$, so $f$ is strictly increasing there and its inverse exists.

From $y=\ln(x-a)$ obtain $x=e^y+a$. Hence $f^{-1}(x)=e^x+a$ with domain $D_{f^{-1}}=R_f=(-\infty,\ln2]$.

Since $f(g(x))=x$, use the inverse: $g(x)=f^{-1}(x)=e^x+a$. The identity holds for $x\leq\ln2$, the range of $f$; beyond that input the displayed formula cannot produce an allowed input of the restricted $f$.

Replace the input in $g(f(x))=x^2$ by $f^{-1}(t)$: $g(t)=g(f(f^{-1}(t)))=(f^{-1}(t))^2=(e^t+a)^2$ for $t\leq\ln2$. The equation fixes $g$ on $R_f$; the printed expression is a valid extension if a larger domain is chosen.

Track $(a,a+2]\xrightarrow{f}(-\infty,\ln2]\xrightarrow{h}R_{hf}$. Plot $y=h(t)$ only for $t\leq\ln2$. Its source-labelled maximum is approximately $(-3.43,8.45)$ and its included endpoint is $(\ln2,-1.48)$, with ordinates rounded. The graph tends to $-\infty$ as $t\to-\infty$, giving $R_{hf}=(-\infty,8.45]$ in the source’s rounded form.

Plot $hf(x)=\frac12[\ln(x-a)-1][\ln(x-a)+1][\ln(x-a)+5]$ on $(a,a+2]$. It tends to $-\infty$ as $x\to a^+$, reaches a maximum approximately at $(a+0.0324,8.45)$ and ends at approximately $(a+2,-1.48)$. This gives the same range.

As an authored precision check, $h\prime(t)=\frac12(3t^2+10t-1)$ has the maximum at $t_*=(-5-2\sqrt7)/3$. Its value is $M=(80+56\sqrt7)/27\approx8.450447164$, attained at $x=a+e^{t_*}$. Thus the exact range is $(-\infty,M]$.