Question:

Step 1: Analyze $f(x) = x^3 - x^2 + x + 1$ on $[0,1]$. Compute $f'(x) = 3x^2 - 2x + 1$. Discriminant $= 4 - 12 = -8 < 0$, so $f'(x) > 0$ for all $x$. Thus $f$ is strictly increasing on $[0,1]$. Step 2: Since $f$ is strictly increasing on $[0,1]$, $\max_{0 \le t \le x} f(t) = f(x)$ for $0 \le x \le 1$. So $g(x) = f(x) = x^3 - x^2 + x + 1$ for $0 \le x \le 1$. Step 3: Check continuity at $x = 1$. $\lim_{x \to 1^-} g(x) = f(1) = 1 - 1 + 1 + 1 = 2$. $\lim_{x \to 1^+} g(x) = 3 - 1 = 2$. $g(1) = f(1) = 2$. So $g$ is continuous at $x = 1$. Hence $g$ is continuous on $[0,2]$. ✓ (option 2) Step 4: Check differentiability at $x = 1$. $g'(1^-) = f'(1) = 3(1)^2 - 2(1) + 1 = 2$. $g'(1^+) = \frac{d}{dx}(3-x)\big|_{x=1} = -1$. Since $g'(1^-) \neq g'(1^+)$, $g$ is not differentiable at $x = 1$. Step 5: For $x \in [0,1)$, $g(x) = f(x)$ is a polynomial, hence differentiable. For $x \in (1,2]$, $g(x) = 3-x$ is differentiable. So $g$ is differentiable on $[0,2] - \{1\}$. ✓ (option 4)