1. **Problem statement:** Determine if the function $$f(x) = \begin{cases} x^2 - 2x + 1, & x \leq -2 \\ \left(\frac{1}{3}\right)^x, & -2 < x \leq 3 \\ \log_3 x, & x > 3 \end{cases}$$ is continuous at $x = -2$ and $x = 3$ using the three conditions of continuity. 2. **Recall the three conditions for continuity at a point $a$:** 1. $f(a)$ is defined. 2. The limit $\lim_{x \to a} f(x)$ exists. 3. $\lim_{x \to a} f(x) = f(a)$. 3. **Check continuity at $x = -2$:** - $f(-2)$ is defined by the first piece: $$f(-2) = (-2)^2 - 2(-2) + 1 = 4 + 4 + 1 = 9$$ - Compute left-hand limit $\lim_{x \to -2^-} f(x)$: Since $x \leq -2$ uses $x^2 - 2x + 1$, $$\lim_{x \to -2^-} f(x) = f(-2) = 9$$ - Compute right-hand limit $\lim_{x \to -2^+} f(x)$: For $-2 < x \leq 3$, $f(x) = \left(\frac{1}{3}\right)^x$, $$\lim_{x \to -2^+} f(x) = \left(\frac{1}{3}\right)^{-2} = 3^2 = 9$$ - Since left and right limits are equal, the limit exists and equals 9. - Check if $\lim_{x \to -2} f(x) = f(-2)$: $$9 = 9$$ **Conclusion:** $f$ is continuous at $x = -2$. 4. **Check continuity at $x = 3$:** - $f(3)$ is defined by the second piece: $$f(3) = \left(\frac{1}{3}\right)^3 = \frac{1}{27}$$ - Compute left-hand limit $\lim_{x \to 3^-} f(x)$: $$\lim_{x \to 3^-} f(x) = \left(\frac{1}{3}\right)^3 = \frac{1}{27}$$ - Compute right-hand limit $\lim_{x \to 3^+} f(x)$: For $x > 3$, $f(x) = \log_3 x$, $$\lim_{x \to 3^+} f(x) = \log_3 3 = 1$$ - Since left and right limits are not equal ($\frac{1}{27} \neq 1$), the limit $\lim_{x \to 3} f(x)$ does not exist. **Conclusion:** $f$ is not continuous at $x = 3$. - The discontinuity is a **jump discontinuity** because the left and right limits exist but are not equal. --- 5. **Problem statement:** Given $$f(x) = x^2 - x$$ and $$f(0) = k,$$ find $k$ such that $f$ is continuous at $x = 0$. 6. **Recall continuity condition at $x=0$:** $$\lim_{x \to 0} f(x) = f(0) = k$$ 7. **Calculate the limit:** $$\lim_{x \to 0} (x^2 - x) = 0^2 - 0 = 0$$ 8. **Set $k$ equal to the limit for continuity:** $$k = 0$$ **Final answers:** - $f$ is continuous at $x = -2$. - $f$ is not continuous at $x = 3$; it has a jump discontinuity there. - The value of $k$ for continuity at $x = 0$ is $0$.