Prove that the function \( f(x) = |x| \) is continuous at \( x = 0 \) but not differentiable.
Step 1: Continuity of the function at \( x = 0 \) is checked by verifying: \[ \lim_{x \to 0^-} |x| = 0 \quad \text{and} \quad \lim_{x \to 0^+} |x| = 0. \] Since both left-hand and right-hand limits exist and are equal to 0, the function is continuous at \( x = 0 \).
Step 2: Differentiability is checked by considering the derivative of \( f(x) = |x| \). For \( x>0 \), \( f'(x) = 1 \), and for \( x<0 \), \( f'(x) = -1 \). At \( x = 0 \), the derivative does not exist because the left-hand derivative is not equal to the right-hand derivative. Thus, the function is not differentiable at \( x = 0 \).
Find the values of \( x, y, z \) if the matrix \( A \) satisfies the equation \( A^T A = I \), where
\[ A = \begin{bmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{bmatrix} \]
(b) Order of the differential equation: $ 5x^3 \frac{d^3y}{dx^3} - 3\left(\frac{dy}{dx}\right)^2 + \left(\frac{d^2y}{dx^2}\right)^4 + y = 0 $