Let \(f(x) = \begin{cases} x^{2}, & x \le 1\\ 2x - 1, & x \gt 1\end{cases}\). Consider the following statements:
(A) f is continuous at \(x = 1\).
(B) f is differentiable at \(x = 1\).
(C) \(f'(2) = 2\)
(D) \(f'(-1) = 2\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (B) and (C) only
At x = 1 both pieces give the same value:
\(1^{2} = 1\), \(2(1) - 1 = 1\)
The left and right derivatives at 1 are the slopes of the two pieces there:
\(Lf'(1) = 2 \times 1 = 2\), \(Rf'(1) = 2\)
So f is continuous and differentiable at 1, f′(x) = 2 for x > 1 and f′(x) = 2x for x < 1.
Checking the statements
(A) f is continuous at \(x = 1\). — Correct. Both pieces give the value 1 at x = 1.
(B) f is differentiable at \(x = 1\). — Correct. The left and right derivatives are both 2.
(C) \(f'(2) = 2\) — Correct. Near x = 2, f(x) = 2x − 1, whose derivative is 2.
(D) \(f'(-1) = 2\) — Incorrect. Near x = −1, f(x) = x2, so f′(−1) = 2 × (−1) = −2.