1. The problem asks to explain why each piecewise function is discontinuous at the given point $a$ and to sketch the graph. 2. For each function, we check continuity at $x=a$ by verifying if $\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a)$. 3. (a) $f(x) = \begin{cases} x+3 & x \leq -1 \\ 2^x & x > -1 \end{cases}$, $a=-1$. - Left limit: $\lim_{x \to -1^-} (x+3) = -1+3=2$. - Right limit: $\lim_{x \to -1^+} 2^x = 2^{-1} = \frac{1}{2}$. - Since $2 \neq \frac{1}{2}$, limits differ, so $f$ is discontinuous at $x=-1$. 4. (b) $f(x) = \begin{cases} \frac{x^2 - x}{x^2 - 1} & x \neq 1 \\ 1 & x=1 \end{cases}$, $a=1$. - Simplify numerator and denominator: $x^2 - x = x(x-1)$, $x^2 -1 = (x-1)(x+1)$. - For $x \neq 1$, $f(x) = \frac{x(x-1)}{(x-1)(x+1)} = \frac{x}{x+1}$. - Left and right limits at $x=1$: $\lim_{x \to 1} f(x) = \frac{1}{1+1} = \frac{1}{2}$. - But $f(1) = 1$, so $f$ is discontinuous at $x=1$. 5. (c) $f(x) = \begin{cases} \cos x & x < 0 \\ 0 & x=0 \\ 1 - x^2 & x > 0 \end{cases}$, $a=0$. - Left limit: $\lim_{x \to 0^-} \cos x = \cos 0 = 1$. - Right limit: $\lim_{x \to 0^+} 1 - x^2 = 1 - 0 = 1$. - $f(0) = 0$. - Since $f(0) \neq$ limits, discontinuous at $x=0$. 6. (d) $f(x) = \begin{cases} \frac{2x^2 - 5x - 3}{x - 3} & x \neq 3 \\ 6 & x=3 \end{cases}$, $a=3$. - Factor numerator: $2x^2 - 5x - 3 = (2x + 1)(x - 3)$. - For $x \neq 3$, $f(x) = \frac{(2x + 1)(x - 3)}{x - 3} = 2x + 1$. - Limit at $x=3$: $\lim_{x \to 3} f(x) = 2(3) + 1 = 7$. - But $f(3) = 6$, so discontinuous at $x=3$. 7. Summary: Each function is discontinuous at $a$ because the limit from the left and right do not equal the function value at $a$. 8. Graphs show jumps or holes at these points.