Question

f(x) = \frac{x-|x-1|}{x}$. Then which of the following is correct

• A. f(x) is discontinuous $x=0$, continuous at $x=1$ and not differentiable at $x=1$
• B. f(x) is discontinuous at $x=0$, continuous and differentiable at $x=1$
• C. f(x) is discontinuous at $x=0$, but differentiable at $x=0$
• D. f(x) is discontinuous at $x=1$ but differentiable at $x=1$

Answer 1

Answer: (A)

f(x) is discontinuous $x=0$, continuous at $x=1$ and not differentiable at $x=1$

Answer 2

The function $f(x) = \frac{x-|x-1|}{x}$ can be defined piecewise: For $x \ge 1$, $|x-1| = x-1$, so $f(x) = \frac{x-(x-1)}{x} = \frac{1}{x}$. For $x < 1$, $|x-1| = -(x-1) = 1-x$, so $f(x) = \frac{x-(1-x)}{x} = \frac{2x-1}{x} = 2 - \frac{1}{x}$. Thus, $f(x) = \begin{cases} 2 - \frac{1}{x} & \text{if } x < 1, x \neq 0 \\ \frac{1}{x} & \text{if } x \ge 1 \end{cases}$

Analysis at $x=0$: The function is not defined at $x=0$ due to division by zero, hence it is discontinuous at $x=0$. $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (2 - \frac{1}{x}) = 2 - (-\infty) = +\infty$. $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (2 - \frac{1}{x}) = 2 - (+\infty) = -\infty$. Since the limits are infinite, $f(x)$ has an essential discontinuity at $x=0$.

Analysis at $x=1$:

• Continuity: $f(1) = \frac{1}{1} = 1$. $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (2 - \frac{1}{x}) = 2 - 1 = 1$. $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} \frac{1}{x} = 1$. Since $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) = 1$, $f(x)$ is continuous at $x=1$.

• Differentiability: For $x < 1$, $f'(x) = \frac{d}{dx}(2 - \frac{1}{x}) = \frac{1}{x^2}$. The left-hand derivative at $x=1$ is $f'_-(1) = \lim_{x \to 1^-} \frac{1}{x^2} = 1$. For $x > 1$, $f'(x) = \frac{d}{dx}(\frac{1}{x}) = -\frac{1}{x^2}$. The right-hand derivative at $x=1$ is $f'_+(1) = \lim_{x \to 1^+} (-\frac{1}{x^2}) = -1$. Since $f'_-(1) \neq f'_+(1)$, $f(x)$ is not differentiable at $x=1$.

Therefore, $f(x)$ is discontinuous at $x=0$, continuous at $x=1$, and not differentiable at $x=1$.