1. **Problem statement:** Given a function $f(x)$ with domain $[-2,5]$ and range $[1,4]$, find the domain and range of the following functions: (a) $g(x) = f(x-3) + 1$ (b) $h(x) = 2f(x)$ (c) $k(x) = -f(2x)$ (d) $m(x) = f(-x) - 2$ (e) Generalize for $Af(Bx - C) + D$ where $f$ has domain $[a,b]$ and range $[c,d]$. --- 2. **Recall:** - The domain of $f(x - h)$ shifts the domain by $+h$. - Adding a constant outside the function shifts the range by that constant. - Multiplying the function by a constant scales the range. - Multiplying the input by a constant scales the domain inversely. - Replacing $x$ by $-x$ reflects the domain about zero. --- 3. **(a) Find domain and range of $g(x) = f(x-3) + 1$** - Domain of $g$: Solve $x-3 \\in [-2,5]$ which means $$-2 \leq x-3 \leq 5$$ Add 3 to all parts: $$-2 + 3 \leq x \leq 5 + 3$$ $$1 \leq x \leq 8$$ - Range of $g$: Since $g(x) = f(x-3) + 1$, add 1 to the range of $f$: $$[1+1, 4+1] = [2,5]$$ --- 4. **(b) Find domain and range of $h(x) = 2f(x)$** - Domain of $h$: Same as $f$, so $[-2,5]$ - Range of $h$: Multiply range of $f$ by 2: $$[2 \times 1, 2 \times 4] = [2,8]$$ --- 5. **(c) Find domain and range of $k(x) = -f(2x)$** - Domain of $k$: Solve $2x \\in [-2,5]$: $$-2 \leq 2x \leq 5$$ Divide all parts by 2: $$\cancel{2} \times \frac{-2}{\cancel{2}} \leq x \leq \frac{5}{2}$$ $$-1 \leq x \leq 2.5$$ - Range of $k$: Since $k(x) = -f(2x)$, multiply range of $f$ by $-1$ and reverse interval: $$[-4, -1]$$ --- 6. **(d) Find domain and range of $m(x) = f(-x) - 2$** - Domain of $m$: Solve $-x \\in [-2,5]$: $$-2 \leq -x \leq 5$$ Multiply all parts by $-1$ and reverse inequalities: $$-5 \leq x \leq 2$$ - Range of $m$: Subtract 2 from range of $f$: $$[1-2, 4-2] = [-1, 2]$$ --- 7. **(e) Generalize for $Af(Bx - C) + D$ with $f$ domain $[a,b]$ and range $[c,d]$** - Domain: Solve $$Bx - C \in [a,b]$$ $$a \leq Bx - C \leq b$$ Add $C$: $$a + C \leq Bx \leq b + C$$ Divide by $B$ (consider sign of $B$): - If $B > 0$: $$\frac{a + C}{B} \leq x \leq \frac{b + C}{B}$$ - If $B < 0$: $$\frac{b + C}{B} \leq x \leq \frac{a + C}{B}$$ - Range: Multiply range by $A$ and add $D$: - If $A > 0$: $$[Ac + D, Ad + D]$$ - If $A < 0$: $$[Ad + D, Ac + D]$$ --- **Final answers:** - (a) Domain: $[1,8]$, Range: $[2,5]$ - (b) Domain: $[-2,5]$, Range: $[2,8]$ - (c) Domain: $[-1,2.5]$, Range: $[-4,-1]$ - (d) Domain: $[-5,2]$, Range: $[-1,2]$ - (e) Domain: $\begin{cases} \left[\frac{a+C}{B}, \frac{b+C}{B}\right], & B>0 \\ \left[\frac{b+C}{B}, \frac{a+C}{B}\right], & B<0 \end{cases}$ Range: $\begin{cases} [Ac+D, Ad+D], & A>0 \\ [Ad+D, Ac+D], & A<0 \end{cases}$