Differentiation

To solve this problem, we need to analyze the function and the given conditions. Let's break this down step by step:

1. We know that for all . This implies that is a convex function in this interval.
2. The function is given by: .
3. We are told that is decreasing in and increasing in . This implies that at .
4. Let's differentiate with respect to :
5. From the condition that at , we get:
6. Given that is convex in the interval (0, 3) (i.e., ), it means that is increasing in this interval.
7. Thus, the point where the derivatives are equal occurs when and are symmetrical with respect to the midpoint of the interval (0, 3), which is 1.5.
8. So, we can set: (since it's symmetric about 1.5), solving this equation gives us:
9. Finally, we are asked for :

Therefore, the value of is 18. The correct answer is 18.

Given and . Therefore, we have:

Evaluating at :
We have:
Using the condition ,we get:

Calculating :
The derivative of is given by:
Therefore:

Substitute into the expression for :

Calculating :
Since and we are asked for :

We are given the function and we need to find the value of .

To solve this, we need to understand that we will apply the rules of differentiation to find and then evaluate it at . We should also simplify the expression wherever possible before differentiating.

**Step 1: Understand the components of the function at .**

• at simplifies to .
• at is .
• at evaluates to .
• at equals .

Plugging these values into the function at , we find that:

.

This confirms that initially the entire expression evaluates to zero.

**Step 2: Differentiate the expression using the product and quotient rules.**

The given function is a quotient of two functions. According to the quotient rule:

Let:

Evaluate derivatives and :

• Since each term in and reduces to a constant or zero at , differentiate directly using the linearity of differentiation. Therefore, each term and product inside at zero simplifies significantly.
• The primary component affecting the derivative is the presence of the non-zero change in as increases from 0.

, and thus the only non-zero influence locally about zero is from the derivative component residing partially from . Therefore, the calculated effect due to simplification leads directly to:

**Conclusion**: The calculated derivative indicates the form , matching the identity of calculated form upon organize elimination of proximate zeroes.

Thus, the answer is .