Find the interval of increase and decrease for the function `g(x) =2f((x^(2))/(2)) +f((27)/(2)-x^(2))` where `f''(x) lt 0 " for all " x in R`

Answer

To find the intervals of increase and decrease for the function \( g(x) = 2f\left(\frac{x^2}{2}\right) + f\left(\frac{27}{2} - x^2\right) \), where \( f''(x) < 0 \) for all \( x \in \mathbb{R} \), we will follow these steps:

### Step 1: Find the derivative \( g'(x) \)

We start by differentiating \( g(x) \):

\[ g'(x) = \frac{d}{dx}\left(2f\left(\frac{x^2}{2}\right)\right) + \frac{d}{dx}\left(f\left(\frac{27}{2} - x^2\right)\right) \]

Using the chain rule, we get:

\[ g'(x) = 2f'\left(\frac{x^2}{2}\right) \cdot \frac{d}{dx}\left(\frac{x^2}{2}\right) + f'\left(\frac{27}{2} - x^2\right) \cdot \frac{d}{dx}\left(\frac{27}{2} - x^2\right) \]

Calculating the derivatives:

\[ \frac{d}{dx}\left(\frac{x^2}{2}\right) = x \quad \text{and} \quad \frac{d}{dx}\left(\frac{27}{2} - x^2\right) = -2x \]

Substituting these into the derivative:

\[ g'(x) = 2f'\left(\frac{x^2}{2}\right) \cdot x - 2x f'\left(\frac{27}{2} - x^2\right) \]

Factoring out \( 2x \):

\[ g'(x) = 2x \left( f'\left(\frac{x^2}{2}\right) - f'\left(\frac{27}{2} - x^2\right) \right) \]

### Step 2: Find critical points

To find the critical points, we set \( g'(x) = 0 \):

\[ 2x \left( f'\left(\frac{x^2}{2}\right) - f'\left(\frac{27}{2} - x^2\right) \right) = 0 \]

This gives us two cases:

1. \( 2x = 0 \) which implies \( x = 0 \)
2. \( f'\left(\frac{x^2}{2}\right) - f'\left(\frac{27}{2} - x^2\right) = 0 \)

### Step 3: Analyze the second case

From the second case, we have:

\[ f'\left(\frac{x^2}{2}\right) = f'\left(\frac{27}{2} - x^2\right) \]

Given that \( f''(x) < 0 \) for all \( x \), we know that \( f'(x) \) is a decreasing function. Therefore, if \( f'\left(a\right) = f'\left(b\right) \) for \( a \neq b \), it implies \( a = b \).

Thus, we can set:

\[ \frac{x^2}{2} = \frac{27}{2} - x^2 \]

Solving this equation:

\[ x^2 + x^2 = 27 \quad \Rightarrow \quad 2x^2 = 27 \quad \Rightarrow \quad x^2 = \frac{27}{2} \quad \Rightarrow \quad x = \pm \sqrt{\frac{27}{2}} = \pm \frac{3\sqrt{3}}{2} \]

### Step 4: Identify critical points

The critical points are:

- \( x = 0 \)
- \( x = -\frac{3\sqrt{3}}{2} \)
- \( x = \frac{3\sqrt{3}}{2} \)

### Step 5: Determine intervals of increase and decrease

We will test the sign of \( g'(x) \) in the intervals formed by these critical points:

1. \( (-\infty, -\frac{3\sqrt{3}}{2}) \)
2. \( (-\frac{3\sqrt{3}}{2}, 0) \)
3. \( (0, \frac{3\sqrt{3}}{2}) \)
4. \( (\frac{3\sqrt{3}}{2}, \infty) \)

#### Interval 1: \( (-\infty, -\frac{3\sqrt{3}}{2}) \)

Choose \( x = -4 \):

\[ g'(-4) = 2(-4) \left( f'\left(\frac{16}{2}\right) - f'\left(\frac{27}{2} - 16\right) \right) = -8 \left( f'(8) - f'(-\frac{7}{2}) \right) \]

Since \( f' \) is decreasing, \( f'(8) < f'(-\frac{7}{2}) \) implies \( g'(-4) > 0 \) (increasing).

#### Interval 2: \( (-\frac{3\sqrt{3}}{2}, 0) \)

Choose \( x = -1 \):

\[ g'(-1) = 2(-1) \left( f'\left(\frac{1}{2}\right) - f'\left(\frac{27}{2} - 1\right) \right) \]

Since \( f' \) is decreasing, \( f'\left(\frac{1}{2}\right) > f'\left(\frac{25}{2}\right) \) implies \( g'(-1) < 0 \) (decreasing).

#### Interval 3: \( (0, \frac{3\sqrt{3}}{2}) \)

Choose \( x = 1 \):

\[ g'(1) = 2(1) \left( f'\left(\frac{1}{2}\right) - f'\left(\frac{25}{2}\right) \right) \]

Since \( f'\left(\frac{1}{2}\right) > f'\left(\frac{25}{2}\right) \), \( g'(1) > 0 \) (increasing).

#### Interval 4: \( (\frac{3\sqrt{3}}{2}, \infty) \)

Choose \( x = 4 \):

\[ g'(4) = 2(4) \left( f'\left(\frac{16}{2}\right) - f'\left(\frac{27}{2} - 16\right) \right) \]

Since \( f'(8) < f'(-\frac{7}{2}) \), \( g'(4) < 0 \) (decreasing).

### Conclusion

The intervals of increase and decrease for \( g(x) \) are:

- **Increasing**: \( (-\infty, -\frac{3\sqrt{3}}{2}) \) and \( (0, \frac{3\sqrt{3}}{2}) \)
- **Decreasing**: \( (-\frac{3\sqrt{3}}{2}, 0) \) and \( (\frac{3\sqrt{3}}{2}, \infty) \)