The function \( f(x) = |x| + |1 - x| \) is the sum of two absolute value functions, which are continuous everywhere. However, absolute value functions are not differentiable at the points where their arguments are zero. Specifically:
Thus, \(f(x)\) is continuous everywhere but differentiable at all points except at \(x = 0\) and \(x = 1\).
Let the function, \(f(x)\) = \(\begin{cases} -3ax^2 - 2, & x < 1 \\a^2 + bx, & x \geq 1 \end{cases}\) Be differentiable for all \( x \in \mathbb{R} \), where \( a > 1 \), \( b \in \mathbb{R} \). If the area of the region enclosed by \( y = f(x) \) and the line \( y = -20 \) is \( \alpha + \beta\sqrt{3} \), where \( \alpha, \beta \in \mathbb{Z} \), then the value of \( \alpha + \beta \) is:
If \( y = e^{{2}\log_e t} \) and \( x = \log_3(e^{t^2}) \), then \( \frac{dy}{dx} \) is equal to: