1. **Problem 1: Analyze the continuity of the piecewise function** Given: $$f(x) = \begin{cases} x^2, & x \leq 1 \\ 3 - x, & 1 < x \leq 2 \\ x, & x > 2 \end{cases}$$ 2. **Recall the definition of continuity at a point $c$:** - $f$ is continuous at $c$ if and only if: $$\lim_{x \to c^-} f(x) = f(c) = \lim_{x \to c^+} f(x)$$ 3. **Check continuity at $x=1$:** - Left limit: $$\lim_{x \to 1^-} f(x) = 1^2 = 1$$ - Right limit: $$\lim_{x \to 1^+} f(x) = 3 - 1 = 2$$ - Function value: $$f(1) = 1^2 = 1$$ - Since $\lim_{x \to 1^-} f(x) = 1 \neq 2 = \lim_{x \to 1^+} f(x)$, there is a jump discontinuity at $x=1$. 4. **Check continuity at $x=2$:** - Left limit: $$\lim_{x \to 2^-} f(x) = 3 - 2 = 1$$ - Right limit: $$\lim_{x \to 2^+} f(x) = 2$$ - Function value: $$f(2) = 3 - 2 = 1$$ - Since $\lim_{x \to 2^-} f(x) = 1 \neq 2 = \lim_{x \to 2^+} f(x)$, there is a jump discontinuity at $x=2$. 5. **For other values of $x$, the function pieces are continuous (polynomial and linear functions).** 6. **Summary:** - Jump discontinuities at $x=1$ and $x=2$. --- 7. **Problem 2: Draw a function with removable, jump, and infinite discontinuities** Consider the function: $$g(x) = \begin{cases} \frac{x^2 - 1}{x - 1}, & x \neq 1 \\ 2, & x = 1 \\ \text{(removable discontinuity at } x=1) \\ 2, & 1 < x \leq 3 \\ 4, & x = 3 \\ \text{(jump discontinuity at } x=3) \\ \frac{1}{x - 5}, & x \neq 5 \\ \text{(infinite discontinuity at } x=5) \end{cases}$$ - At $x=1$, $g(x)$ simplifies to $x+1$ for $x \neq 1$, but $g(1)=2$ creates a removable discontinuity. - At $x=3$, the function jumps from 2 (left) to 4 (value at 3), a jump discontinuity. - At $x=5$, $g(x) = \frac{1}{x-5}$ has a vertical asymptote, an infinite discontinuity. --- **Final answers:** - The given $f(x)$ has jump discontinuities at $x=1$ and $x=2$. - The example $g(x)$ above has removable, jump, and infinite discontinuities as requested.