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 f(x) is

Options

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

Answer

To determine whether the function \( f(x) \) is differentiable at \( x = 0 \), we will calculate the right-hand derivative and the left-hand derivative at that point.

### Step 1: Define the function
The function is given by:

\[ f(x) = \begin{cases} \frac{x(e^{1/x} - e^{-1/x})}{e^{1/x} + e^{-1/x}} & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \]

### Step 2: Calculate the right-hand derivative at \( x = 0 \)
The right-hand derivative is defined as:

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

Substituting \( f(0) = 0 \):

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

For \( h \neq 0 \):

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

Thus,

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

### Step 3: Simplify the expression
As \( h \to 0^+ \), \( e^{1/h} \to \infty \) and \( e^{-1/h} \to 0 \):

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

### Step 4: Calculate the left-hand derivative at \( x = 0 \)
The left-hand derivative is defined as:

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

Substituting \( f(0) = 0 \):

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

For \( h < 0 \):

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

Thus,

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

### Step 5: Simplify the left-hand derivative
As \( h \to 0^- \), \( e^{1/h} \to 0 \) and \( e^{-1/h} \to \infty \):

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

### Step 6: Compare the derivatives
We have:

\[ f'_+(0) = 1 \quad \text{and} \quad f'_-(0) = -1 \]

Since the right-hand derivative and left-hand derivative are not equal:

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

### Conclusion
The function \( f(x) \) is not differentiable at \( x = 0 \).

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