1. **State the problem:** We have a piecewise function defined as $$f(x) = \begin{cases} x & \text{for } 0 \leq x \leq 1 \\ 2x - 2 & \text{for } 1 < x \leq 2 \end{cases}$$ We need to determine if $f$ is continuous over the interval $[0,1]$. 2. **Recall the definition of continuity on an interval:** A function $f$ is continuous on $[0,1]$ if it is continuous at every point in that interval, including the endpoints. 3. **Check continuity at the endpoints:** - At $x=0$, since $f(x) = x$ for $0 \leq x \leq 1$, the function is continuous from the right at 0. - At $x=1$, we must check if the left-hand limit, right-hand limit, and function value agree. 4. **Evaluate limits and function value at $x=1$:** - Left-hand limit: $$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x = 1$$ - Right-hand limit: $$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2x - 2) = 2(1) - 2 = 0$$ - Function value at 1: $$f(1) = 1$$ 5. **Analyze continuity at $x=1$:** Since the left-hand limit ($1$) and the function value ($1$) agree, but the right-hand limit ($0$) does not equal these, the function is not continuous at $x=1$. 6. **Conclusion:** The function $f$ is continuous on $[0,1]$ except at $x=1$ where it is discontinuous from the right. **Answer:** b. No, since $f$ is continuous on the right at 0 but not on the left at 1.