Question:

Step 1: Analyze the parametric equations for $t \geq 0$. When $t \geq 0$, we have $|t| = t$. Substituting into the given parametric equations: $$x = 2t - |t| = 2t - t = t$$ $$y = t^2 + t|t| = t^2 + t \cdot t = 2t^2$$ Eliminating the parameter $t$, we get: $$y = 2x^2 \text{ for } x \geq 0$$ Step 2: Analyze the parametric equations for $t < 0$. When $t < 0$, we have $|t| = -t$. Substituting into the given parametric equations: $$x = 2t - |t| = 2t - (-t) = 3t$$ $$y = t^2 + t|t| = t^2 + t(-t) = t^2 - t^2 = 0$$ Since $x = 3t$ and $t < 0$, we have $x < 0$. Therefore: $$y = 0 \text{ for } x < 0$$ Step 3: Write the explicit form of $f(x)$. Combining the results from Steps 1 and 2, the function $f(x)$ is defined as: $$f(x) = \begin{cases} 0 & \text{if } x < 0 \\ 2x^2 & \text{if } x \geq 0 \end{cases}$$ Step 4: Check continuity at $x = 0$. We evaluate the function value and the limits at $x = 0$: - Function value: $f(0) = 2(0)^2 = 0$ - Left limit: $\lim_{x \to 0^-} f(x) = 0$ - Right limit: $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} 2x^2 = 0$ Since $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = 0$, the function is continuous at $x = 0$. Step 5: Check differentiability at $x = 0$ using left and right derivatives. Left derivative at $x = 0$: $$f'_-(0) = \lim_{h \to 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^-} \frac{0 - 0}{h} = 0$$ Right derivative at $x = 0$: $$f'_+(0) = \lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^+} \frac{2h^2 - 0}{h} = \lim_{h \to 0^+} 2h = 0$$ Since $f'_-(0) = f'_+(0) = 0$, the function is differentiable at $x = 0$. Step 6: State the conclusion. The function $f(x)$ is both continuous and differentiable at $x = 0$. **Final Answer:** The correct answer is **Option 1**: $f(x)$ is both continuous and differentiable at $x = 0$. Note: The provided correct answer of "4" appears to be inconsistent with the mathematical analysis. Based on the rigorous calculations above, Option 1 is the mathematically correct answer.