Differential Calculus

We are given a functional equation for a twice differentiable function and a condition on its first derivative at . We need to find the value of the expression .

Concept Used:

The solution involves simplifying the given functional equation using trigonometric identities to determine the form of the function . The key trigonometric identities are:

Once the form of is established, we use differentiation and the given initial condition to find the specific function. Finally, we compute its second derivative at the required point.

Step-by-Step Solution:

The given functional equation is:

We start by rearranging the terms to group and .

Expanding both sides, we get:

Now, we bring terms with to one side and terms with to the other side:

Using the sine addition and subtraction formulas, this simplifies to:

For values of where and , we can write:

Let and .

The equation becomes:

Since this relation holds for arbitrary and , it implies that the expression must be a constant, let's call it .

To find the form of , we can substitute , which gives . Thus, the function is of the form:

Next, we use the given condition to find the constant .

First, we differentiate with respect to :

Now, we substitute :

Given , we have:

So, the specific function is . Now we need to find its second derivative.

Final Computation & Result:

We need to calculate the value of . First, let's evaluate :

We know that .

Substituting this value:

Finally, we compute the required expression:

Hence, the value of is -3.

The problem asks for the value of for the function (with ), given that its local maximum and minimum occur at points and respectively, and that these points are related by the condition .

Concept Used:

To find the local maximum and minimum of a differentiable function, we use the following steps:

1. First Derivative Test: The local extrema of a function occur at its critical points, which are the points where the first derivative is zero or undefined. For a polynomial function, we find the roots of .

2. Second Derivative Test: To classify the critical points, we use the second derivative. Let be a critical point such that .

• If , then the function has a local maximum at .
• If , then the function has a local minimum at .

Step-by-Step Solution:

The given function is .

First, we find the first derivative of to locate the critical points:

To find the critical points, we set :

We can simplify this quadratic equation by dividing the entire equation by 6:

This equation can be factored by finding two numbers that multiply to and add up to . These numbers are and .

The critical points are and . These are the values of and .

To determine which point corresponds to the maximum and which to the minimum, we use the second derivative test. We find the second derivative, :

Now, we evaluate at each critical point:

At :

Since it is given that , we have . Therefore, the function has a local maximum at . This means .

At :

Since , we have . Therefore, the function has a local minimum at . This means .

We are given the condition . Substituting our values for and :

This gives two possible solutions: or . Since the problem states that , we must have .

Now that we have the value of , we can write the specific function for :

Final Computation & Result:

We are asked to find the value of . We substitute into the function:

Hence, the value of is 37.