Understanding Monotonicity of g(x)
To determine the monotonicity of the function \( g(x) = 2f\left(\frac{x^2}{2}\right) + f(6 - x^2) \), we start by finding its derivative \( g'(x) \).

Finding the Derivative
Using the chain rule, we compute:
\[
g'(x) = 2f'\left(\frac{x^2}{2}\right) \cdot \frac{d}{dx}\left(\frac{x^2}{2}\right) + f'(6 - x^2) \cdot \frac{d}{dx}(6 - x^2)
\]
This simplifies to:
\[
g'(x) = 2f'\left(\frac{x^2}{2}\right) \cdot x + f'(6 - x^2) \cdot (-2x)
\]
Putting it all together, we get:
\[
g'(x) = 2x f'\left(\frac{x^2}{2}\right) - 2x f'(6 - x^2)
\]

Analyzing the Sign of g'(x)
To analyze the sign of \( g'(x) \):
- **Critical Points**: \( g'(x) = 0 \) when \( 2x f'\left(\frac{x^2}{2}\right) = 2x f'(6 - x^2) \). This occurs when \( x = 0 \) or when the derivatives are equal.
- **Intervals**: Check the sign of \( g'(x) \) in the intervals \( (-\infty, 0) \) and \( (0, \infty) \):
- For \( x > 0 \), since \( x > 0 \), the term \( 2x \) is positive. The behavior of \( g'(x) \) depends on \( f' \).
- For \( x < 0="" \),="" \(="" 2x="" \)="" is="" negative,="" which="" implies="" \(="" g'(x)="" \)="" could="" be="" negative="" or="" positive="" depending="" on="" \(="" f'="" />

Conclusion
Given that \( f''(x) > 0 \), \( f' \) is strictly increasing. The sign of \( g'(x) \) will dictate where \( g(x) \) is increasing or decreasing.
- **Increasing Intervals**: \( g(x) \) is increasing where \( f'\left(\frac{x^2}{2}\right) > f'(6 - x^2) \).
- **Decreasing Intervals**: \( g(x) \) decreases where \( f'\left(\frac{x^2}{2}\right) < f'(6="" -="" x^2)="" />
Thus, further analysis of \( f' \) will help specify these intervals precisely.

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