Question

If \(f(x) = \begin{cases} x, & 0 \le x \le 1 \\ 2x - 1, & 1<x \end{cases}\), then

• A. f is discontinuous at \(x = 1\)
• B. f is differentiable at \(x = 1\)
• C. f is continuous but not differentiable at \(x = 1\)
• D. None of the above

Hint:

Function continuous if left and right limits equal; differentiable if left and right derivatives equal.

Answer 1

The Correct Option is C

To determine the continuity and differentiability of the function \(f(x)\) at \(x=1\), let us analyze each condition given:

The function is defined as:

\(f(x) = \begin{cases} x, & 0 \le x \le 1 \\ 2x - 1, & x > 1 \end{cases}\)

1. Test for Continuity at \(x=1\):
   • For the function to be continuous at \(x=1\), the left-hand limit, right-hand limit, and the function value at that point should all be equal.
   • Left-hand limit (\(\lim_{x \to 1^-} f(x)\)): \(\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x = 1\)
   • Right-hand limit (\(\lim_{x \to 1^+} f(x)\)): \(\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2x - 1) = 1\)
   • Function value at \(x=1\): \(f(1) = 1\)
   • Since the left-hand limit, right-hand limit, and the actual value of the function at \(x=1\) are equal, \(f(x)\) is continuous at \(x=1\).
2. Test for Differentiability at \(x=1\):
   • For the function to be differentiable at \(x=1\), the left-hand derivative and the right-hand derivative must be equal.
   • Left-hand derivative (\(\lim_{h \to 0^-} \frac{f(1+h) - f(1)}{h}\)): \(\lim_{h \to 0^-} \frac{(1+h) - 1}{h} = \lim_{h \to 0^-} \frac{h}{h} = 1\)
   • Right-hand derivative (\(\lim_{h \to 0^+} \frac{f(1+h) - f(1)}{h}\)): \(\lim_{h \to 0^+} \frac{(2(1+h) - 1) - 1}{h} = \lim_{h \to 0^+} \frac{2h}{h} = 2\)
   • Since the left-hand derivative (\(1\)) is not equal to the right-hand derivative (\(2\)), the function \(f(x)\) is not differentiable at \(x=1\).

Conclusion:

• The function \(f(x)\) is continuous at \(x=1\) but not differentiable at \(x=1\).

Therefore, the correct answer is: f is continuous but not differentiable at \(x = 1\).