1. **State the problem:** We need to sketch a function $f$ with the following limit and value conditions: - $\lim_{x \to 2^+} f(x) = 3$ - $\lim_{x \to 2^-} f(x) = 5$ - $f(2) = 4$ - $\lim_{x \to -3^+} f(x) = -2$ - $\lim_{x \to -3^-} f(x) = 1$ - $f(-3) = -4$ - $\lim_{x \to 1^+} f(x) = \infty$ - $\lim_{x \to 0^-} f(x) = -\infty$ - $\lim_{x \to \infty} f(x) = 2$ - $\lim_{x \to -\infty} f(x) = 2$ 2. **Explain the meaning of limits and function values:** - The right-hand limit at $x=2$ is 3, left-hand limit is 5, but $f(2)=4$ means the function has a jump discontinuity at $x=2$. - Similarly, at $x=-3$, the right limit is $-2$, left limit is $1$, and $f(-3)=-4$, another jump discontinuity. - Vertical asymptotes at $x=1$ (function goes to $+\infty$ from the right) and at $x=0$ (function goes to $-\infty$ from the left). - Horizontal asymptotes at $y=2$ as $x \to \pm \infty$. 3. **Sketching the function:** - Near $x=2$, draw the graph approaching $5$ from the left (open circle at $(2,5)$), approaching $3$ from the right (open circle at $(2,3)$), and a filled dot at $(2,4)$. - Near $x=-3$, draw the graph approaching $1$ from the left (open circle at $(-3,1)$), approaching $-2$ from the right (open circle at $(-3,-2)$), and a filled dot at $(-3,-4)$. - At $x=1$, draw a vertical asymptote with the graph going to $+\infty$ as $x \to 1^+$. - At $x=0$, draw a vertical asymptote with the graph going to $-\infty$ as $x \to 0^-$. - As $x \to \pm \infty$, the graph approaches the horizontal line $y=2$. 4. **Is $f$ a function?** - Yes, $f$ is a function because for every $x$ in the domain, there is exactly one $f(x)$ value defined. - Even though limits from left and right differ at $x=2$ and $x=-3$, the function values $f(2)=4$ and $f(-3)=-4$ are single unique values. **Final answer:** The function $f$ has jump discontinuities at $x=2$ and $x=-3$, vertical asymptotes at $x=0$ and $x=1$, and horizontal asymptotes at $y=2$ as $x \to \pm \infty$. It is a function because it assigns exactly one output for each input.