By the Mean Value Theorem (MVT), if a function  is continuous and differentiable on , then there exists at least one value  such that . , a polynomial, is continuous and differentiable everywhere; setting , it follows from the MVT that there is  such that 

Evaluating  and :

The expression for  is equal to 

,

the correct choice.

By Rolle's Theorem, if  is continuous on  and differentiable on , and , then there must be  such that . 

 is given to be continuous. Also, if we set , we note that . This sets up the conditions for Rolle's Theorem to apply. As a consequence, there must be  such that .

Incidentally, it does follow from the given information that  must have a zero on the interval , but this is due to the Intermediate Value Theorem, not Rolle's Theorem.

To find the mean value of a function over some interval , one mus use the formula: .

Plugging in 

Simplifying

One must then use the inverse Sine function to find the value c:

Unless we are explicitly told so, via graph, information, or otherwise, we cannot assume  is continuous at  unless , which is required for  to be continuous at .

We cannot assume anything about the existence of , because we do not know what  is, or its end behavior.

The limit of a function as  approaches a value  exists if and only if the limit from the left is equal to the limit from the right; the actual value of  is irrelevant. Since the function is piecewise-defined, we can determine whether these limits are equal by finding the limits of the individual expressions. These are both polynomials, so the limits can be calculated using straightforward substitution:

 does not exist, because .

For a function to be continuous the following criteria must be met:

1.  The function must exist at the point (no division by zero, asymptotic behavior, negative logs, or negative radicals). 
2. The limit must exist.
3. The point must equal the limit. (Symbolically, ).

It is easiest to first find any points where the function is undefined. Since our function involves a fraction and a natural log, we must find all points in the domain such that the natural log is less than or equal to zero, or points where the denominator is equal to zero.

To find the values that cause the natural log to be negative we set 

Therefore, those x values will yield our points of discontinuity. Normally, we would find values where the natural log is negative; however, for all  the function is positive. 

Differentiate both sides and proceed with the product rule: 

                                                                                      (1)

Evaluate the derivatives in each term. For the first term,  

                                                (2)

 apply the chain rule, 

So now the first term in equation (2) can be written, 

                            (3)

The second term in equation (2) is easy, this is just the product of  multiplied by the derivative of , 

                                                               (4)

Combine equations (3) and (4) to write the derivative, 

Use the product rule to find the derivative. 

Use the power rule to find the derivative.

Thus, the derivative is 

Here we use the product rule: 

Let  and 

Then  (using the chain rule)

and  (using the chain rule)

Subbing these values back into our equation gives us

Simplify by combining like-terms

and pulling out a  from each term gives our final answer