1. **Problem Statement:** Sketch a graph of a function $f$ such that: - $\lim_{x \to 0} f(x) = 4$ - $\lim_{x \to 8^-} f(x) = 1$ - $\lim_{x \to 8^+} f(x) = -3$ - $f(0) = 6$ - $f(8) = -1$ 2. **Understanding Limits and Function Values:** - The limit $\lim_{x \to a} f(x)$ is the value $f(x)$ approaches as $x$ gets close to $a$ from both sides. - The left-hand limit $\lim_{x \to a^-} f(x)$ is the value $f(x)$ approaches as $x$ approaches $a$ from the left. - The right-hand limit $\lim_{x \to a^+} f(x)$ is the value $f(x)$ approaches as $x$ approaches $a$ from the right. - The function value $f(a)$ can be different from the limits. 3. **Interpreting the Conditions:** - At $x=0$, the limit is 4 but $f(0)=6$, so there is a jump discontinuity at 0. - At $x=8$, the left limit is 1, the right limit is -3, and $f(8)=-1$, so there is a jump discontinuity with different left and right limits and a function value different from both. 4. **Sketching the Graph:** - For $x$ near 0, the function approaches 4 from both sides but the point at $x=0$ is at 6. - For $x$ near 8, the function approaches 1 from the left and -3 from the right, with the point at $x=8$ at -1. - Between these points, the function can be continuous or piecewise defined to meet these conditions. 5. **Summary:** - The graph has a jump at $x=0$ with $f(0)=6$ but limit 4. - The graph has a jump at $x=8$ with left limit 1, right limit -3, and $f(8)=-1$. This satisfies all given conditions.