Question

Consider \[ f(x) = \begin{cases} x^2, & x < 0, \\[6pt] x, & x \ge 0. \end{cases} \] Which statements are correct?

• A. The function is continuous for all \(x\)
• B. The derivative of the function is discontinuous at \(x=0\)
• C. The derivative of the function is continuous at \(x=1\)
• D. The function is discontinuous at \(x=0\)

Hint:

For piecewise functions, first check the glue point(s).

• Continuity: requires matching function values at the glue point(s).
• Differentiability: requires matching slopes (one-sided derivatives) at the glue point(s).

Answer 1

The Correct Option is A, B, C

Step 1: Continuity.
Each branch (\(x^2\) for \(x < 0\), \(x\) for \(x\ge0\)) is continuous on its domain. At \(x=0\): \(\lim_{x\to0^-} f(x) = 0^2 = 0\), \(\lim_{x\to0^+} f(x) = 0\), and \(f(0)=0\). So \(f\) is continuous at \(0\) and hence continuous for all \(x\). \(\Rightarrow\) (A) True, (D) False. 

Step 2: Differentiability at \(x=0\).
For \(x < 0\), \(f'(x)=2x \;\Rightarrow\; \lim_{x\to0^-} f'(x)=0\). For \(x > 0\), \(f'(x)=1 \;\Rightarrow\; \lim_{x\to0^+} f'(x)=1\). One–sided derivatives at \(0\) are unequal \(\;\Rightarrow\; f\) is not differentiable at \(x=0\). \(\Rightarrow\) (B) True. 

Step 3: Continuity of derivative at \(x=1\).
Near \(x=1 > 0\), the active branch is \(f(x)=x\) on both sides; thus \(f'(x)=1\) in a neighborhood of \(1\). Therefore \(f'\) exists and is continuous at \(x=1\). \(\Rightarrow\) (C) True. \[\boxed{\text{Correct: (A), (B), (C)}} \]