Solution. (a). The expression inside changes sign where \(3-2x=0\), that is at \(x=\frac {3}{2}\). So \[h(x)=\begin {cases} 3-2x, & x\leq \tfrac {3}{2}\\[2pt] 2x-3, & x>\tfrac {3}{2} \end {cases}\]

(b). Two straight lines meeting at \(\left (\frac {3}{2},0\right )\): one of gradient \(-2\) coming down to that point, one of gradient \(2\) rising from it. The graph is the usual V, with its vertex on the \(x\)-axis at \(x=\frac {3}{2}\) and \(y\)-intercept \(h(0)=3\).

Equivalently, sketch \(y=3-2x\) and reflect the part below the \(x\)-axis upwards, as in the general method of this chapter.

(c). \(D_h=\mathbb {R}\), since nothing can go wrong for any input, and \(R_h=[0,\infty )\), since an absolute value is never negative and every non-negative value is attained.

Solution. (a). The equation says \(\left |3x-2\right |=\left |x-2\right |\). Two absolute values are equal exactly when the quantities inside are equal or opposite: \[3x-2=x-2 \quad \text {or}\quad 3x-2=-(x-2).\] The first gives \(2x=0\), so \(x=0\). The second gives \(3x-2=-x+2\), hence \(4x=4\) and \(x=1\). Both check: at \(x=0\) both sides are \(2\); at \(x=1\) both are \(1\). So \(x\in \{0,1\}\).

(b). Move everything to one side rather than cross-multiplying: \[\frac {1}{x-1}-\frac {2}{x+1}\geq 0 \implies \frac {(x+1)-2(x-1)}{(x-1)(x+1)}\geq 0 \implies \frac {3-x}{(x-1)(x+1)}\geq 0 .\] Critical values are \(x=-1\), \(x=1\) and \(x=3\). Testing the sign on each interval:

The expression is positive on \((-\infty ,-1)\) and \((1,3)\), and zero at \(x=3\), which the \(\geq \) admits. The values \(\pm 1\) are excluded, the expression being undefined there. Hence \[(-\infty ,-1)\cup (1,3] .\]