Given $f(x) = \begin{cases} 1 + x & x < 0 \\ 2 - 3x & x \geq 0 \end{cases}$,find the critical point $x = \dots \dots$.

At the point $x = 1$,the given function $f(x) = \begin{cases} x^3 - 1; & 1 < x < \infty \\ x - 1; & -\infty < x \le 1 \end{cases}$ is

If $\alpha$ and $\beta$ are such that the function $f(x)$ defined by $f(x) = \begin{cases} \alpha x^2 - \beta, & |x| < 1 \\ \frac{-1}{|x|}, & |x| \ge 1 \end{cases}$ is differentiable everywhere,then the ordered pair $(\alpha, \beta) =$

Let $g: [-2, 2] \rightarrow R$ and $f: [-2, 2] \rightarrow R$ be two functions defined as $g(x) = \begin{cases} -1, & \text{if } -2 \le x < 0 \\ x^2 - 1, & \text{if } 0 \le x \le 2 \end{cases}$ and $f(x) = |g(x)| + g(|x|) + 2$. In the interval $(-2, 2)$, $f$ is not differentiable at $x = $

If $f(x) = \begin{cases} x^2 & \text{if } x \leqslant x_0 \\ ax + b & \text{if } x > x_0 \end{cases}$ is derivable for all $x \in \mathbb{R}$,then the values of $a$ and $b$ are respectively:

If $f(x)$ is a differentiable function such that $f: R \to R$ and $f\left( \frac{1}{n} \right) = 0$ for all $n \ge 1, n \in I$,then: