Let `g: R -> R` be a differentiable function with `g(0) = 0,,g'(1)!=0`.Let `f(x)={x/|x|g(x), 0 !=0 and 0,x=0` and `h(x)=e^(|x|)` for all `x in R`. Let `(foh)(x)` denote `f(h(x)) and (hof)(x)` denote `h(f(x))`. Then which of the following is (are) true?

A. f is differentiable at x = 0
B. h is differentiable at x = 0
C. f o h is differentiable at x = 0
D. h o f is differentiable at x = 0

Options

A. f is differentiable at x = 0
B. h is differentiable at x = 0
C. f o h is differentiable at x = 0
D. h o f is differentiable at x = 0

Answer

To solve the problem step by step, we will analyze the differentiability of the functions \( f \), \( h \), \( f \circ h \), and \( h \circ f \) at \( x = 0 \).

### Step 1: Analyze the function \( f(x) \)

The function \( f(x) \) is defined as:
\[ f(x) = \begin{cases} \frac{x}{|x|} g(x) & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \]

For \( x > 0 \), \( f(x) = g(x) \) and for \( x < 0 \), \( f(x) = -g(x) \).

### Step 2: Check differentiability of \( f \) at \( x = 0 \)

To check if \( f \) is differentiable at \( x = 0 \), we need to compute the left-hand derivative (LHD) and right-hand derivative (RHD):

- **Left-hand derivative**:
\[ f'_{-}(0) = \lim_{h \to 0^-} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^-} \frac{-g(h)}{h} \]
Since \( g(0) = 0 \) and \( g'(1) \neq 0 \), we can use L'Hôpital's Rule if needed.

- **Right-hand derivative**:
\[ f'_{+}(0) = \lim_{h \to 0^+} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^+} \frac{g(h)}{h} \]

Both derivatives will yield \( g'(0) \) if \( g(x) \) is continuous at \( x = 0 \). Since \( g(0) = 0 \), we find that:
\[ f'_{-}(0) = -g'(0) \quad \text{and} \quad f'_{+}(0) = g'(0) \]
Thus, \( f'(0) \) exists if \( g'(0) = 0 \).

### Conclusion for \( f \):
Since both left-hand and right-hand derivatives exist and are equal, \( f \) is differentiable at \( x = 0 \).

### Step 3: Analyze the function \( h(x) \)

The function \( h(x) \) is defined as:
\[ h(x) = e^{|x|} \]

### Step 4: Check differentiability of \( h \) at \( x = 0 \)

- **Left-hand derivative**:
\[ h'_{-}(0) = \lim_{h \to 0^-} \frac{h(h) - h(0)}{h} = \lim_{h \to 0^-} \frac{e^{-h} - 1}{h} = -1 \]

- **Right-hand derivative**:
\[ h'_{+}(0) = \lim_{h \to 0^+} \frac{h(h) - h(0)}{h} = \lim_{h \to 0^+} \frac{e^{h} - 1}{h} = 1 \]

Since the left-hand and right-hand derivatives are not equal, \( h \) is not differentiable at \( x = 0 \).

### Conclusion for \( h \):
\( h \) is not differentiable at \( x = 0 \).

### Step 5: Analyze \( f \circ h \)

We need to compute \( f(h(x)) \):
\[ f(h(x)) = \begin{cases} g(e^{x}) & \text{if } x > 0 \\ -g(e^{-x}) & \text{if } x < 0 \\ g(1) & \text{if } x = 0 \end{cases} \]

### Step 6: Check differentiability of \( f \circ h \) at \( x = 0 \)

- **Left-hand derivative**:
\[ (f \circ h)'_{-}(0) = \lim_{h \to 0^-} \frac{-g(e^{-h}) - g(1)}{-h} \]

- **Right-hand derivative**:
\[ (f \circ h)'_{+}(0) = \lim_{h \to 0^+} \frac{g(e^{h}) - g(1)}{h} \]

Since \( g(1) \) is constant, both derivatives will yield \( g'(1) \). Thus, \( f \circ h \) is differentiable at \( x = 0 \).

### Conclusion for \( f \circ h \):
\( f \circ h \) is differentiable at \( x = 0 \).

### Step 7: Analyze \( h \circ f \)

We need to compute \( h(f(x)) \):
\[ h(f(x)) = \begin{cases} e^{g(x)} & \text{if } x > 0 \\ e^{-g(x)} & \text{if } x < 0 \\ 1 & \text{if } x = 0 \end{cases} \]

### Step 8: Check differentiability of \( h \circ f \) at \( x = 0 \)

- **Left-hand derivative**:
\[ (h \circ f)'_{-}(0) = \lim_{h \to 0^-} \frac{e^{-g(h)} - 1}{h} \]

- **Right-hand derivative**:
\[ (h \circ f)'_{+}(0) = \lim_{h \to 0^+} \frac{e^{g(h)} - 1}{h} \]

Both derivatives will yield \( g'(0) \) if \( g \) is continuous at \( 0 \). Since \( g(0) = 0 \), we find that both derivatives exist and are equal.

### Conclusion for \( h \circ f \):
\( h \circ f \) is differentiable at \( x = 0 \).

### Final Answers:
- A. True: \( f \) is differentiable at \( x = 0 \)
- B. False: \( h \) is not differentiable at \( x = 0 \)
- C. True: \( f \circ h \) is differentiable at \( x = 0 \)
- D. True: \( h \circ f \) is differentiable at \( x = 0 \)