An absolute-value restriction and a composite range

The source graph shows the horizontal line y=3.5 meeting f three times. Its labelled points are (2,3), (3,4), (5,0) and (6,5). Hence f is not one-to-one and has no inverse on the full stated domain.

The longest invertible interval is [3,5]. The graph is strictly increasing on [2,3], decreasing on [3,5], then increasing on [5,6]; the middle interval has greatest length. On [3,5], the quadratic inside the absolute value is nonpositive, so $y=-(x^2-6x+5)=4-(x-3)^2$. Thus $(x-3)^2=4-y$ and $x=3\pm\sqrt{4-y}$. The branch x≥3 gives $f^{-1}:x\mapsto3+\sqrt{4-x}$ for $x\in\mathbb R$, $0\leq x\leq4$. Its domain is the restricted range [0,4].

The graph of h has positive values, approaches y=0 as x→−∞, and includes the endpoint (3/2,2). Hence $R_h=(0,2]$. The outer domain is $D_g=(-\infty,0)\cup(0,\infty)$. Since $R_h\subseteq D_g$, gh exists on the full stated domain of h.

First map through h, then restrict g to $R_h=(0,2]$. The unrestricted graph of g=ln(x²) has two branches with asymptote x=0. On the retained positive branch, g increases from −∞ (not attained) to $g(2)=\ln4$. Thus $R_{gh}=(-\infty,\ln4]=(-\infty,2\ln2]$.

Write $gh(x)=\ln[(1/(2-x))^2]$ for x≤3/2. The composite graph is increasing, tends to −∞ as x→−∞, and has included endpoint (3/2,ln4). Its range is again $(−\infty,\ln4]$. The two source graph methods are both retained in the figure.