To find relative maximums, we need to find where our first derivative changes sign. To do this, find your first derivative and then find where it is equal to zero.

Begin with:

    at 

This means we have extrema at x=0 and x=-8/3

Because we are only concerned about the interval from -5 to 0, we only need to test points on that interval. Try points between our two extrema, as well as one between -8/3 and -5.

So our first derivative goes from negative to positive at -8/3, thus this is the x coordinate of our relative maximum on this interval.

Though there are relative maxima at  and  (found by setting the first derivative of the function equal to zero and solving for x.)  the maximum value along the whole interval is actually at the upper endpoint, when .  When looking for extrema along an interval, looking for zeros of the first derivative does not account for endpoint extrema.

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: