We are asked to determine where the function \( f(x) = \sin(|x|) - |x| \) is not differentiable.
Step 1: Understanding the Structure of the Function The function involves the absolute value of \( x \), which is a piecewise function. To understand where this function is non-differentiable, we need to check the point where \( |x| \) changes behavior — that is at \( x = 0 \).
Step 2: Check the Differentiability at \( x = 0 \) We know that the absolute value function \( |x| \) is defined as: \[ |x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases} \] Thus, the function becomes: - For \( x \geq 0 \), \( f(x) = \sin(x) - x \). - For \( x < 0 \), \( f(x) = \sin(-x) + x \). Now, calculate the derivatives on either side of \( x = 0 \): - For \( x > 0 \), \( f'(x) = \cos(x) - 1 \). - For \( x < 0 \), \( f'(x) = -\cos(x) + 1 \). At \( x = 0 \), the left-hand and right-hand derivatives do not match, so the function is not differentiable at \( x = 0 \). Thus, the correct answer is \( x = 0 \).
Let $\alpha,\beta\in\mathbb{R}$ be such that the function \[ f(x)= \begin{cases} 2\alpha(x^2-2)+2\beta x, & x<1 \\ (\alpha+3)x+(\alpha-\beta), & x\ge1 \end{cases} \] is differentiable at all $x\in\mathbb{R}$. Then $34(\alpha+\beta)$ is equal to}