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Question: A function is defined as $f(x) = \begin{cases} e^{-x}+1 &, x \leq 0 \\ |x-1| &, x > 0 \end{cases}$. ...

A function is defined as f(x)={e−x+1,x≤0∣x−1∣,x>0f(x) = \begin{cases} e^{-x}+1 &, x \leq 0 \\ |x-1| &, x > 0 \end{cases}. Which of the following statement is FALSE?

A

f(x)f(x) is continuous at x=1x = 1

B

f(x)f(x) is differentiable at x=0x = 0

C

f(x)f(x) is non-differentiable at x=1x = 1

D

L.H.D. of f(x)f(x) at x=0x = 0 is -1

Answer

f(x)f(x) is differentiable at x=0x = 0

Explanation

Solution

The function is defined as: f(x)={e−x+1,x≤0∣x−1∣,x>0f(x) = \begin{cases} e^{-x}+1 &, x \leq 0 \\ |x-1| &, x > 0 \end{cases}

We need to check each statement to identify the FALSE one.

1. f(x)f(x) is continuous at x=1x = 1

For x>0x > 0, f(x)=∣x−1∣f(x) = |x-1|.

  • f(1)=∣1−1∣=0f(1) = |1-1| = 0.
  • lim⁡x→1+f(x)=lim⁡x→1+∣x−1∣=∣1−1∣=0\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} |x-1| = |1-1| = 0.
  • lim⁡x→1−f(x)=lim⁡x→1−∣x−1∣=∣1−1∣=0\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} |x-1| = |1-1| = 0.

Since f(1)=lim⁡x→1+f(x)=lim⁡x→1−f(x)=0f(1) = \lim_{x \to 1^+} f(x) = \lim_{x \to 1^-} f(x) = 0, the function is continuous at x=1x=1. Statement 1 is TRUE.

2. f(x)f(x) is differentiable at x=0x = 0

First, let's check for continuity at x=0x=0.

  • f(0)=e−0+1=1+1=2f(0) = e^{-0} + 1 = 1 + 1 = 2.
  • Left-hand limit (LHL) at x=0x=0:
    lim⁡x→0−f(x)=lim⁡x→0−(e−x+1)=e0+1=2\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (e^{-x}+1) = e^0+1 = 2.
  • Right-hand limit (RHL) at x=0x=0:
    lim⁡x→0+f(x)=lim⁡x→0+∣x−1∣\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} |x-1|. As x→0+x \to 0^+, xx is slightly greater than 0, so x−1x-1 is negative. Thus, ∣x−1∣=−(x−1)=1−x|x-1| = -(x-1) = 1-x.
    lim⁡x→0+(1−x)=1−0=1\lim_{x \to 0^+} (1-x) = 1-0 = 1.

Since LHL (22) ≠\neq RHL (11), the function is not continuous at x=0x=0. A function must be continuous at a point to be differentiable at that point. Since f(x)f(x) is not continuous at x=0x=0, it cannot be differentiable at x=0x=0. Statement 2 is FALSE.

3. f(x)f(x) is non-differentiable at x=1x = 1

For x>0x > 0, f(x)=∣x−1∣f(x) = |x-1|. We know that the function ∣g(x)∣|g(x)| is generally not differentiable at points where g(x)=0g(x)=0. Here g(x)=x−1g(x) = x-1, which is 0 at x=1x=1. Let's calculate the L.H.D. and R.H.D. at x=1x=1:

  • Right-hand derivative (R.H.D.) at x=1x=1:
    f′(1+)=lim⁡h→0+f(1+h)−f(1)h=lim⁡h→0+∣(1+h)−1∣−∣1−1∣h=lim⁡h→0+∣h∣−0hf'(1^+) = \lim_{h \to 0^+} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0^+} \frac{|(1+h)-1| - |1-1|}{h} = \lim_{h \to 0^+} \frac{|h| - 0}{h}. Since h>0h > 0, ∣h∣=h|h|=h.
    f′(1+)=lim⁡h→0+hh=1f'(1^+) = \lim_{h \to 0^+} \frac{h}{h} = 1.
  • Left-hand derivative (L.H.D.) at x=1x=1:
    f′(1−)=lim⁡h→0−f(1+h)−f(1)h=lim⁡h→0−∣(1+h)−1∣−∣1−1∣h=lim⁡h→0−∣h∣−0hf'(1^-) = \lim_{h \to 0^-} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0^-} \frac{|(1+h)-1| - |1-1|}{h} = \lim_{h \to 0^-} \frac{|h| - 0}{h}. Since h<0h < 0, ∣h∣=−h|h|=-h.
    f′(1−)=lim⁡h→0−−hh=−1f'(1^-) = \lim_{h \to 0^-} \frac{-h}{h} = -1.

Since L.H.D. (−1-1) ≠\neq R.H.D. (11), the function is not differentiable at x=1x=1. Statement 3 is TRUE.

4. L.H.D. of f(x)f(x) at x=0x = 0 is -1

For x≤0x \leq 0, f(x)=e−x+1f(x) = e^{-x}+1. L.H.D. at x=0x=0:
f′(0−)=lim⁡h→0−f(0+h)−f(0)hf'(0^-) = \lim_{h \to 0^-} \frac{f(0+h) - f(0)}{h}. Since h→0−h \to 0^-, hh is negative, so 0+h≤00+h \leq 0.
f(0+h)=e−(0+h)+1=e−h+1f(0+h) = e^{-(0+h)}+1 = e^{-h}+1.
f(0)=e−0+1=2f(0) = e^{-0}+1 = 2.
f′(0−)=lim⁡h→0−(e−h+1)−2h=lim⁡h→0−e−h−1hf'(0^-) = \lim_{h \to 0^-} \frac{(e^{-h}+1) - 2}{h} = \lim_{h \to 0^-} \frac{e^{-h}-1}{h}. Let y=−hy = -h. As h→0−h \to 0^-, y→0+y \to 0^+.
f′(0−)=lim⁡y→0+ey−1−y=−lim⁡y→0+ey−1yf'(0^-) = \lim_{y \to 0^+} \frac{e^y-1}{-y} = -\lim_{y \to 0^+} \frac{e^y-1}{y}. Using the standard limit lim⁡x→0ex−1x=1\lim_{x \to 0} \frac{e^x-1}{x} = 1:
f′(0−)=−1f'(0^-) = -1. Statement 4 is TRUE.

The question asks for the FALSE statement. Based on our analysis, statement 2 is FALSE.

The final answer is f(x) is differentiable at x=0\boxed{f(x) \text{ is differentiable at } x = 0}.