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:

A point of inflection can be the point at which the derivative is equal to zero, a local minimum or maximum, and a point of inflection.  A critical point will always be one that the tangent line to the original function is either completely horizontal or vertical.  When the tangent line is completely horizontal, this will be a local maximum or minimum for the function.  When the tangent line is completely vertical, then this will be a point of inflection.  A point of inflection is a point in which there is a change of curvature in the graph of the function.

To find whether a function is decreasing or increasing along an interval, we look at the critical values and use what we call the first derivative test.  Take the example .  The derivative would be .  To find the critical value we set the derivative equal to zero and solve for .

Now we have our critical point .  So we choose a number greater than this and plug it back into our derivative function.

The solution is positive.  This means that the interval  on the graph of 

,

The solution is decreasing.  This means that the interval  is decreasing on the graph of .

We begin by finding the derivative of our function, .

Now we set our derivative function equal to zero and solve for .

And we are left with two critical points 

Critical point , increasing: , decreasing: .

Explanation: We first have to find the critical point(s).  To do this we begin by finding the derivative.

Now to find the critical point(s), we set the derivative equal to zero and solve for .

And so our critical point is .  Now we will choose a number greater than  to plug back into our derivative.  We will arbitrarily choose the number .

.

The solution is positive, so we know that on the interval  our function  is increasing.  Now we must choose a number less than to plug back into our derivative.  We will arbitrarily choose .

The solution is negative.  So we know that on the interval  our function  is decreasing.

We begin by finding the derivative of the function.

Now we must find the critical points of the function

So we need to look at the intervals  , , , and .  We will begin with the interval .  We will arbitrarily choose  to plug back into our derivative function.

The solution is negative, so the interval  is decreasing.  Now we will look at the interval .  We need to choose a number between  and .  We will choose .

The solution is negative.  So from the interval , the function is increasing.  Now we will look at the interval .  We will choose the number to plug back into our equation.

The solution is negative, so the interval  is decreasing.  Now we will look at the interval .  We will choose the number  to plug back into our derivative function.

The solution is positive.  So the interval  is increasing.

The increasing intervals are  and 

Think back to the types of critical values that we can have.  We can have a maximum or a minimum; if we have one of these then the intervals will change from increasing to decreasing or vice versa.  Or we will have an inflection point.  An inflection point is a change in curvature of the function.  So if the critical point is an inflection point, then the function can be completely increasing or completely decreasing.

Begin by finding the derivative.  

Now find the critical point of the function.

Now we will plug a number that is greater than  into the derivative function.  We will arbitrarily choose the number .

The solution is positive, so the interval  is increasing.