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 \( x = 0 \).
\[ f(x) = \begin{cases} x^2 + 2, & \text{if } x \neq 0 \\ 1, & \text{if } x = 0 \end{cases} \]
is not continuous at \( x = 0 \).Is the function \( f(x) \) defined by
\[ f(x) = \begin{cases} x + 5, & \text{if } x \leq 1 \\ x - 5, & \text{if } x > 1 \end{cases} \]
continuous at \( x = 1 \)?
(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 $