If `f(x) = {{:(x((e^(1//x) - e^(-1//x))/(e^(1//x)+e^(1//x)))",",x ne 0),(" "0",",x = 0):}`, then at `x = 0`, then `f(x)` is

Options

A. differentiable
B. not differentiable
C. `f'(0^(+))=-1`
D. `f'(0^(-))=1`

Answer

To determine the differentiability of the function \( f(x) \) at \( x = 0 \), we need to analyze the function given by:

\[ f(x) = \begin{cases} \frac{x \left( e^{\frac{1}{x}} - e^{-\frac{1}{x}} \right)}{e^{\frac{1}{x}} + e^{-\frac{1}{x}}}, & x \neq 0 \\ 0, & x = 0 \end{cases} \]

### Step 1: Find \( f(0) \)
Since the function is defined as \( f(0) = 0 \), we have:

\[ f(0) = 0 \]

### Step 2: Find the left-hand derivative \( f'(0^-) \)
The left-hand derivative is defined as:

\[ f'(0^-) = \lim_{h \to 0^-} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0^-} \frac{f(h)}{h} \]

For \( h < 0 \), we substitute into the function:

\[ f(h) = \frac{h \left( e^{\frac{1}{h}} - e^{-\frac{1}{h}} \right)}{e^{\frac{1}{h}} + e^{-\frac{1}{h}}} \]

Thus, we have:

\[ f'(0^-) = \lim_{h \to 0^-} \frac{1}{h} \cdot \frac{h \left( e^{\frac{1}{h}} - e^{-\frac{1}{h}} \right)}{e^{\frac{1}{h}} + e^{-\frac{1}{h}}} \]

This simplifies to:

\[ f'(0^-) = \lim_{h \to 0^-} \frac{e^{\frac{1}{h}} - e^{-\frac{1}{h}}}{e^{\frac{1}{h}} + e^{-\frac{1}{h}}} \]

As \( h \to 0^- \), \( e^{\frac{1}{h}} \to 0 \) and \( e^{-\frac{1}{h}} \to \infty \). Therefore, we can simplify:

\[ f'(0^-) = \lim_{h \to 0^-} \frac{0 - \infty}{0 + \infty} = \lim_{h \to 0^-} -1 = -1 \]

### Step 3: Find the right-hand derivative \( f'(0^+) \)
The right-hand derivative is defined as:

\[ f'(0^+) = \lim_{h \to 0^+} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0^+} \frac{f(h)}{h} \]

For \( h > 0 \), we substitute into the function:

\[ f(h) = \frac{h \left( e^{\frac{1}{h}} - e^{-\frac{1}{h}} \right)}{e^{\frac{1}{h}} + e^{-\frac{1}{h}}} \]

Thus, we have:

\[ f'(0^+) = \lim_{h \to 0^+} \frac{1}{h} \cdot \frac{h \left( e^{\frac{1}{h}} - e^{-\frac{1}{h}} \right)}{e^{\frac{1}{h}} + e^{-\frac{1}{h}}} \]

This simplifies to:

\[ f'(0^+) = \lim_{h \to 0^+} \frac{e^{\frac{1}{h}} - e^{-\frac{1}{h}}}{e^{\frac{1}{h}} + e^{-\frac{1}{h}}} \]

As \( h \to 0^+ \), \( e^{\frac{1}{h}} \to \infty \) and \( e^{-\frac{1}{h}} \to 0 \). Therefore, we can simplify:

\[ f'(0^+) = \lim_{h \to 0^+} \frac{\infty - 0}{\infty + 0} = \lim_{h \to 0^+} 1 = 1 \]

### Step 4: Conclusion
Since \( f'(0^-) = -1 \) and \( f'(0^+) = 1 \), we find that:

\[ f'(0^-) \neq f'(0^+) \]

Thus, \( f(x) \) is not differentiable at \( x = 0 \).

### Final Answer
Therefore, at \( x = 0 \), \( f(x) \) is **not differentiable**.

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