`h(x)=3f((x^2)/3)+f(3-x^2)AAx in (-3, 4)` where `f''(x)> 0 AA x in (-3,4),` then `h(x)` is

Options

A. a) increasing in `((3)/(2),4)`
B. b) increasing in `(-(3)/(2),0)`
C. c) decreasing in `(-3,-(3)/(2))`
D. d) decreasing in `(0,(3)/(2))`

Answer

To solve the problem, we need to analyze the function \( h(x) = 3f\left(\frac{x^2}{3}\right) + f(3 - x^2) \) given that \( f''(x) > 0 \) for \( x \in (-3, 4) \). This condition implies that \( f'(x) \) is a strictly increasing function.

### Step 1: Find the derivative of \( h(x) \)

To determine the monotonicity of \( h(x) \), we first need to compute its derivative \( h'(x) \):

\[ h'(x) = \frac{d}{dx}\left(3f\left(\frac{x^2}{3}\right)\right) + \frac{d}{dx}\left(f(3 - x^2)\right) \]

Using the chain rule, we have:

\[ h'(x) = 3 \cdot f'\left(\frac{x^2}{3}\right) \cdot \frac{d}{dx}\left(\frac{x^2}{3}\right) + f'(3 - x^2) \cdot \frac{d}{dx}(3 - x^2) \]

Calculating the derivatives:

\[ \frac{d}{dx}\left(\frac{x^2}{3}\right) = \frac{2x}{3} \]
\[ \frac{d}{dx}(3 - x^2) = -2x \]

Substituting these back into the expression for \( h'(x) \):

\[ h'(x) = 3 \cdot f'\left(\frac{x^2}{3}\right) \cdot \frac{2x}{3} - 2x \cdot f'(3 - x^2) \]

This simplifies to:

\[ h'(x) = 2x \left(f'\left(\frac{x^2}{3}\right) - f'(3 - x^2)\right) \]

### Step 2: Analyze the sign of \( h'(x) \)

Since \( f''(x) > 0 \), \( f'(x) \) is strictly increasing. Therefore, we can analyze the expression \( f'\left(\frac{x^2}{3}\right) - f'(3 - x^2) \):

- If \( x > 0 \), then \( \frac{x^2}{3} < 3 - x^2 \) for \( x^2 < 9 \) (which is true for \( x < 3 \)). Thus, \( f'\left(\frac{x^2}{3}\right) < f'(3 - x^2) \) leading to \( h'(x) < 0 \).
- If \( x < 0 \), then \( \frac{x^2}{3} > 3 - x^2 \) for \( x^2 < 9 \). Thus, \( f'\left(\frac{x^2}{3}\right) > f'(3 - x^2) \) leading to \( h'(x) > 0 \).

### Step 3: Determine intervals of increase and decrease

From the analysis:

- \( h(x) \) is increasing for \( x < 0 \).
- \( h(x) \) is decreasing for \( x > 0 \).

### Step 4: Identify critical points and behavior at endpoints

The critical point occurs at \( x = 0 \):

- As \( x \) approaches \( 0 \) from the left, \( h(x) \) increases.
- As \( x \) approaches \( 0 \) from the right, \( h(x) \) decreases.

Thus, \( h(x) \) has a maximum at \( x = 0 \).

### Conclusion

The function \( h(x) \) is:

- Increasing on the interval \( (-3, 0) \).
- Decreasing on the interval \( (0, 4) \).

### Final Answer

Thus, \( h(x) \) has a maximum at \( x = 0 \) and is increasing on \( (-3, 0) \) and decreasing on \( (0, 4) \).

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