Question

Which of the following statements is true for the function \[ f(x) = \begin{cases} x^2 + 3, & x \neq 0, \\ 1, & x = 0? \end{cases} \]

• A. \( f(x) \) is continuous and differentiable \( \forall x \in \mathbb{R} \)
• B. \( f(x) \) is continuous \( \forall x \in \mathbb{R} \)
• C. \( f(x) \) is continuous and differentiable \( \forall x \in \mathbb{R} \setminus \{0\} \)
• D. \( f(x) \) is discontinuous at infinitely many points

Hint:

To check continuity, equate \( \lim_{x \to a} f(x) \) with \( f(a) \). To check differentiability, verify the existence of \( f'(x) \) in the neighborhood of the point.

Answer 1

The Correct Option is C

The function is defined as: \[ f(x) = \begin{cases} x^2 + 3, & x eq 0, \\1, & x = 0?\end{cases}\]

Step 1: Assess continuity at \( x = 0 \) 
Continuity at \( x = 0 \) requires \( \lim_{x \to 0} f(x) = f(0) \). We calculate: \[ \lim_{x \to 0} f(x) = \lim_{x \to 0} (x^2 + 3) = 3, \quad \text{and} \quad f(0) = 1. \] As \( \lim_{x \to 0} f(x) eq f(0) \), the function is discontinuous at \( x = 0 \). 

Step 2: Examine differentiability for \( x eq 0 \) 
For \( x eq 0 \), \( f(x) = x^2 + 3 \). This is a polynomial, which is differentiable for all real numbers. Therefore, \( f(x) \) is differentiable for all \( x eq 0 \). 

Step 3: Confirm continuity and differentiability at other points 
Since \( f(x) = x^2 + 3 \) for \( x eq 0 \), the function is continuous and differentiable for all \( x \in \mathbb{R} \setminus \{0\} \). Consequently, \( f(x) \) is continuous and differentiable everywhere except at \( x = 0 \).