Question

Consider the function \(f(x)\) $= \begin{cases}x^{3} & \text { if } x<0 \\\ x^{2} & \text { if } 0 \leq x<1 \\ 2 x-1 & \text { if } 1 \leq x<2 \\\ x^{2}-2 x+3 & \text { if } x \geq 2\end{cases}$ then \(\mathrm{f}\) is continuous and differentiable for (A) \(x \in R\) (B) \(x \in R-\\{0,2\\}\) (C) \(x \in \mathbb{R}-\\{2\\}\) (D) \(x \in \mathbb{R}-\\{1,2\\}\)

Short answer

Answer: The function \(f(x)\) is continuous and differentiable for all \(x \in \mathbb{R}\).

Step-by-step solution

Identify critical points

The critical points of the function are the boundaries where the different pieces of the function are defined. For this function, the critical points are \(x = 0, \ 1,\) and \(2\).

Check continuity at critical points

For the function to be continuous at the critical points, the limit must exist and be equal to the function's value at the point. Now, let's check the continuity of \(f(x)\) at each critical point: At \(x = 0\): \(\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} x^3 = 0\) \(\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x^2 = 0\) \(f(0) = 0^2 = 0\) Since the left and right limits and the function value at \(x=0\) are equal, \(f(x)\) is continuous at \(x=0\). At \(x = 1\): \(\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x^2 = 1\) \(\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2x - 1) = 1\) \(f(1) = 2(1) - 1 = 1\) Since the left and right limits and the function value at \(x=1\) are equal, \(f(x)\) is continuous at \(x=1\). At \(x = 2\): \(\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (2x - 1) = 3\) \(\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (x^2 - 2x + 3) = 3\) \(f(2) = 2^2 - 2(2) + 3= 3\) Since the left and right limits and the function value at \(x=2\) are equal, \(f(x)\) is continuous at \(x=2\). Therefore, the function is continuous for all \(x \in \mathbb{R}\).

Check differentiability at critical points

For the function to be differentiable at the critical points, the derivatives from the left and right must exist and be equal. We'll find the derivative of each piece and check for differentiability at the critical points. Derivative of \(x^3\) is \(3x^2\). Derivative of \(x^2\) is \(2x\). Derivative of \((2x - 1)\) is \(2\). Derivative of \((x^2 - 2x + 3)\) is \(2x - 2\). Now, let's check the differentiability at each critical point: At \(x = 0\): Left derivative: \(3(0)^2 = 0\) Right derivative: \(2(0) = 0\) Left and right derivatives are equal, so \(f(x)\) is differentiable at \(x=0\). At \(x = 1\): Left derivative: \(2(1) = 2\) Right derivative: \(2\) Left and right derivatives are equal, so \(f(x)\) is differentiable at \(x=1\). At \(x = 2\): Left derivative: \(2\) Right derivative: \(2(2) - 2 = 2\) Left and right derivatives are equal, so \(f(x)\) is differentiable at \(x=2\). Since \(f(x)\) is continuous and differentiable at all critical points and in each of the intervals, the function is continuous and differentiable for \(x \in \mathbb{R}\).

Choose the correct option

Based on the analysis, the correct option is (A) \(x \in \mathbb{R}\)