To determine the relative maxima for the function, we must determine where the first derivative of the function changes from positive to negative.

The first derivative of the function is

and was found using the following rule:

Next, we must find the critical points, the values at which the first derivative is equal to zero:

Using the critical values, we can then create the intervals on which we evaluate the sign of the first derivative:

To check the sign of the first derivative, plug in any value on each interval into the first derivative function. On the first interval, the first derivative is positive, on the second, it is negative, and on the third it is positive. The first derivative changes from positive to negative at  so a relative maximum exists here.

We notice that this function has 2 extrema which are located at  and . We could have also of found this by looking at the function itself: 

.  

We know that extrema exist when the slope of the function is zero, hence we take the derivative, set it equal to zero, and than solve for x.

The derivative is the following: 

.  

Therefore setting both pieces equal to zero we see the following: 

 and .  

Finding out if one of these x values produce a local min or max however requires either the first derivative test, the second derivative test, or analyzing the graph. We see that for a small neighborhood around ,  is the largest term, hence it is a local max.  Similarly, for a small neighborhood around ,  is the smallest term, hence  is a local min.  

Therefore when  is the only spot in which a local max occurs. Remember that it is not the x value that must be the largest, rather it is it's corresponding y value.

The mean value theorem states that if continuous   has , then for all , there must be an  that maps to , so since the value above go from negative to positive twice, zero must be mapped to at least twice. 

We notice that this function has 2 extrema which are located at x=-1 and x=0.  We could have also of found this by looking at the function itself: 

.  

We know that extrema exist when the slope of the function is zero, hence we take the derivative, set it equal to zero, and than solve for x.

The derivative is the following: 

.  

Therefore setting both pieces equal to zero we see the following: 

 and .  

Finding out if one of these x values produce a local min or max however requires either the first derivative test, the second derivative test, or analyzing the graph. We see that for a small neighborhood around ,  is the largest term, hence it is a local max. Similarly, for a small neighborhood around ,  is the smallest term, hence  is a local min and the only local min of this function. Remember that it is not the x value that must be the smallest, rather it is it's corresponding y value.

To determine the length of a curve between two points, we evaluate the integral

There are many reasons this works, but we'll give an informal explanation here:

If we divide this curve into three line-segments, we can see that they become more and more similar to the original curve. By adding up all the little hypotenuses, we can arrive at the length of the curve. Note, that

If we think of the integration symbol as a sum of infinitely small parts, this gets us the formula for length:

.

Returning to the problem, we plug the derivative into the length formula:

substitute:

Simplify:

By trigonometric identities we get:

The distance between two points can be easily found using the Distance Formula:

Applying the points we are given to this formula results in:

This is one of the answer choices.

First, take the derivative of f(x). This is very easy:

f'(x) = 12x² – 4x

Now, the slope of the tangent line at x = 5 is f'(5). Evaluated, this is: f'(5) = 12 * 5 * 5 – 4 * 5 = 300 – 20 = 280

Now, we must find the intersection point on the original line:

f(5) = 4 * 5³ – 2 * 5² + 4 = 500 – 50 + 4 = 454

Therefore, the point of tangent intersection is (5,454).

Using the point-slope form of linear equations, we can find the line:

(y – 454) = 280(x – 5)

y – 454 = 280x – 1400

y = 280x – 946

The equation of the tangent line will have the form , where  is the slope of the line and .  

To find the slope, we need to evaluate the derivative at :

Now that we have the slope, we can determine the equation of the tangent line using the point-slope formula:

To find the equation of the line at that point, you need two things: the slope at that point and the y adjustment of the function, in the form , where m is the slope and  is the y adjustment. To get the slope, find the derivative of  and plug in the desired point  for , giving us an answer of  for the slope.

To find the y adjustment pick a point 0 in the original function. For simplicity, let's plug in , which gives us a y of 1, so an easy point is . Next plug in those values into the equation of a line, . The new equation with all parameters plugged in is

Now you simply solve for , which is .

Final equation of the line tangent to  at  is 

To get the slope, find the derivative of  and plug in the desired point  for , giving us an answer of  for the slope.

Remember that the derivative of .

To find the  adjustment pick a point  (for example) in the original  function. For simplicity, let's plug in , which gives us a  of , so an easy point is . Next plug in those values into the equation of a line, . The new equation with all parameters plugged in is

The coefficient in front of the  is the slope.

Now you simply solve for , which is .

Final equation of the line tangent to  at is .