1. **State the problem:** We are given a piecewise function: $$f(x) = \begin{cases} -\frac{1}{3}(x + 4)^2 + 3 & x \leq -1 \\ 3|x - 2| - 4 & -1 < x < 4 \\ (x - 6)^3 + 3 & x \geq 4 \end{cases}$$ We need to: A. Find the domain and range. B. Identify discontinuities and classify their types. 2. **Domain:** The domain is all $x$ values for which $f(x)$ is defined. Each piece covers: - $x \leq -1$ - $-1 < x < 4$ - $x \geq 4$ Together, these cover all real numbers, so: $$\text{Domain} = (-\infty, \infty)$$ 3. **Range:** Analyze each piece: - For $x \leq -1$, $f(x) = -\frac{1}{3}(x + 4)^2 + 3$ is a downward parabola with vertex at $x = -4$: $$f(-4) = -\frac{1}{3}(0)^2 + 3 = 3$$ As $x$ moves away from $-4$, $(x+4)^2$ increases, so $f(x)$ decreases without bound. So range for this piece is $(-\infty, 3]$. - For $-1 < x < 4$, $f(x) = 3|x - 2| - 4$. The absolute value has a minimum at $x=2$: $$f(2) = 3|0| - 4 = -4$$ At the endpoints: $$f(-1) = 3| -1 - 2| - 4 = 3 \times 3 - 4 = 5$$ $$f(4) = 3|4 - 2| - 4 = 3 \times 2 - 4 = 2$$ Since $x$ is strictly between $-1$ and $4$, these endpoint values are not included. Range for this piece is $(-4, 5)$. - For $x \geq 4$, $f(x) = (x - 6)^3 + 3$ is a cubic function. At $x=4$: $$f(4) = (4 - 6)^3 + 3 = (-2)^3 + 3 = -8 + 3 = -5$$ As $x \to \infty$, $f(x) \to \infty$. So range for this piece is $[-5, \infty)$. 4. **Combine ranges:** Piece 1: $(-\infty, 3]$ Piece 2: $(-4, 5)$ Piece 3: $[-5, \infty)$ Check overlaps: - Piece 1 and 2 overlap between $(-4, 3]$. - Piece 2 and 3 overlap between $[-5, 5)$ and $[-5, \infty)$, but note piece 3 starts at $-5$. Overall range is: $$(-\infty, 5) \cup [-5, \infty) = (-\infty, \infty)$$ Because piece 3 covers from $-5$ to infinity and piece 1 covers down to $-\infty$, the entire real line is covered. 5. **Discontinuities:** Check at boundaries $x = -1$ and $x = 4$. - At $x = -1$: Left limit: $$f(-1^-) = -\frac{1}{3}(-1 + 4)^2 + 3 = -\frac{1}{3}(3)^2 + 3 = -3 + 3 = 0$$ Right limit: $$f(-1^+) = 3| -1 - 2| - 4 = 3 \times 3 - 4 = 5$$ Since left and right limits differ, there is a jump discontinuity at $x = -1$. - At $x = 4$: Left limit: $$f(4^-) = 3|4 - 2| - 4 = 3 \times 2 - 4 = 2$$ Right limit: $$f(4^+) = (4 - 6)^3 + 3 = -8 + 3 = -5$$ Left and right limits differ, so jump discontinuity at $x = 4$. 6. **Summary:** - Domain: $(-\infty, \infty)$ - Range: $(-\infty, \infty)$ - Discontinuities: jump discontinuities at $x = -1$ and $x = 4$.