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 \( x = 1 \)?
Step 1: To determine continuity at \( x = 1 \), we check if the left-hand limit, right-hand limit, and the function value at \( x = 1 \) are equal.
Step 2: The left-hand limit as \( x \to 1^- \) is: \[ \lim_{x \to 1^-} f(x) = 1 + 5 = 6 \]
Step 3: The right-hand limit as \( x \to 1^+ \) is: \[ \lim_{x \to 1^+} f(x) = 1 - 5 = -4 \]
Step 4: Since the left-hand limit and the right-hand limit are not equal, the function is not continuous at \( x = 1 \).
Prove that the function \( f(x) = |x| \) is continuous at \( x = 0 \) but not differentiable.
\[ f(x) = \begin{cases} x^2 + 2, & \text{if } x \neq 0 \\ 1, & \text{if } x = 0 \end{cases} \]
is not continuous at \( x = 0 \).(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 $