1. **State the problem:** We need to find the points where the piecewise function $$f(x) = \begin{cases} x + 8 & x < 0 \\ e^x & 0 \leq x \leq 1 \\ 2 - x & x > 1 \end{cases}$$ is discontinuous, and determine if it is continuous from the right, left, or neither at those points. 2. **Identify potential discontinuities:** The function changes definition at $x=0$ and $x=1$, so check continuity at these points. 3. **Check continuity at $x=0$:** - Left-hand limit: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (x+8) = 0 + 8 = 8$ - Right-hand limit: $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} e^x = e^0 = 1$ - Function value: $f(0) = e^0 = 1$ Since left limit $8 \neq$ right limit $1$, the limit does not exist at $x=0$, so $f$ is discontinuous at $x=0$. Check one-sided continuity: - Continuous from the right if $\lim_{x \to 0^+} f(x) = f(0)$: here $1=1$, so yes. - Continuous from the left if $\lim_{x \to 0^-} f(x) = f(0)$: here $8 \neq 1$, so no. 4. **Check continuity at $x=1$:** - Left-hand limit: $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} e^x = e^1 = e$ - Right-hand limit: $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2 - x) = 2 - 1 = 1$ - Function value: $f(1) = e^1 = e$ Since left limit $e \neq$ right limit $1$, the limit does not exist at $x=1$, so $f$ is discontinuous at $x=1$. Check one-sided continuity: - Continuous from the left if $\lim_{x \to 1^-} f(x) = f(1)$: here $e = e$, so yes. - Continuous from the right if $\lim_{x \to 1^+} f(x) = f(1)$: here $1 \neq e$, so no. 5. **Summary:** - At $x=0$, $f$ is discontinuous, continuous from the right only. - At $x=1$, $f$ is discontinuous, continuous from the left only. 6. **Graph description:** - For $x<0$, $f(x) = x+8$ is a line with slope 1 and y-intercept 8. - For $0 \leq x \leq 1$, $f(x) = e^x$ is an increasing exponential curve from $1$ to $e \approx 2.718$. - For $x>1$, $f(x) = 2 - x$ is a line with slope $-1$ starting at $1$ when $x=1$ and decreasing. This matches the given graph description with jumps at $x=0$ and $x=1$. Final answers: - Discontinuities at $x=0$ and $x=1$. - At $x=0$, continuous from the right only. - At $x=1$, continuous from the left only.