In order to solve this we must visualize a triangle that has formed between the tip of the radio tower and the plane.  To find the distance the plane is from the tip of the tower, we must use the Pythagorean theorem  where  is the horizontal distance from the tower, in this case 300; and  is the vertical distance from the tip of the tower, in this case 400.  Therefore the distance the plane is away from the tip of the tower is 500 feet.  Now in order to find the horizontal speed of the airplane, we must take the derivative of the Pythagorean theorem with respect to time in order to find the change in horizontal distance of the plane with respect to time.  

Using the power rule 

,

we find that the theorem becomes 

.  

We need to find  and we know the variables  we simply plug in to find the answer.

This parabola would have the formula . When the first derivative is positive, the function is parabola is increasing. The first derivative is , which is positive on the domain . When the second derivative is positive, the function is concave up. The second derivative is , which is always positive for all real values of .  

Therefore, this function is,

Concave up over  and increasing over .

In order to solve this question, we must realize that the the derivative of a graph equation becomes the slope equation and the derivative of the slope equation becomes the concavity equation.  Therefore in order to solve this problem we must take the double derivative of the graph equation, set it equal to zero, find the inflection points and determine the concavity from there.

In order to take the derivative of equation, the power rule must be applied, .   The first derivative becomes  and the second derivative .  

Setting it equal to zero, we find that the there is not way to set the second derivative to zero with a real number.  Plugging in a random number, 0 in this case, we find that the second derivative is positive.  This means that since there is no inflection point and the second derivative is positive, the graph is concave up on all intervals.

In order to solve this question, we must realize that the the derivative of a graph equation becomes the slope equation and the derivative of the slope equation becomes the concavity equation.  Therefore in order to solve this problem we must take the double derivative of the graph equation, set it equal to zero, find the inflection points and determine the concavity from there.

In order to take the derivative of equation, the power rule must be applied, .   The first derivative becomes  and the second derivative .  

Setting it equal to zero, we find that the inflection point on this graph is .  The inflection point on a graph is the point in which the concavity of the graph changes, therefore the concavity will only change at that point and no other point. In order to determine the concavity, we will plug in numbers before and after this point.  By plugging those numbers into the second derivative, if the value is negative, the graph is concave down and if the value is positive, the graph is concave up.  For this problem, we will plug in .

For , we find that the second derivative is positive.

For , we find that the second derivative is negative.

This means that as  approaches , the graph is concave up and as  leaves  the graph is concave down.

Therefore the graph is concave down from .

If a function's rate of increasing is decreasing, what we are saying is that the first derivative of the function (i.e its rate of change) is decreasing. This must mean that the second derivative is less than . If  over an interval, we can say that the function is concave down over that interval. Recall that the sign of the second derivative describes the concavity, while the sign of the first derivative describes whether a function is increasing or decreasing.

The function is concave down when the second derivative is negative.

The second derivative, in this case, is negative for all values of except for when it is undefined, namely at .  The only value in our interval, , that satisfies this is , so at this point only the function is not concave down.  It is, in fact, not even defined at this point.

To determine the interval(s) over which the function is concave down, we must find the intervals over which the second derivative is negative. To make the intervals, we must find the point(s) at which the second derivative of the function is equal to zero. (At these points, if the sign of the second derivative changes, we have a point of inflection.)

We must first find the first and second derivative of the function:

We used the following rule to find the derivatives:

Now, find the point at which the second derivative is equal to zero:

This point sets the bounds for the intervals we are going to look at:.

Over the first interval, the second derivative is negative, and over the second, the second derivative is positive. Thus, our answer is the first interval, 

To determine the intervals on which the function is concave down, we must find the intervals on which the second derivative is negative. 

The second derivative is found from the first derivative:

The derivatives were found using the following rule:

The second derivative is a constant, 4, which is positive, so there are no intervals on the entire domain on which the function is concave down. 

To find intervals of concavity, we need to find the inflection points of f. To do this, we need to take the second derivative of  and find the values of  for which it is :

so....  

These are the two inflection points of . In order to find the intervals where  is concave down, we need to find the intervals where  is negative (this being true because a negative second derivative means the signed rate of change of the derivative is negative, which is true when we have a steepening negative rate of change or a leveling off positive rate of change for , consistent with an upside down bowl). Since  is continuous, we can find this out by taking test values in the three intervals created by the two inflection points.

Convenient values are ,  and ; one for each interval:

 = 

so  is concave down on the interval given in the answer. The reason there isn't a larger or different interval is:

    so concave up

and

 so concave up

In order to find the intervals of concavity, we must take the second derivative of the function  and find the inflection points by setting the setting it to zero (this tells allows us to see when the rate of the rate of change is changing from negative to positive or vice versa).  The second derivative is , allowing us to see that our inflection points are .  By testing values in between the three different intervals, we can find out which interval is concave down.  The interval  is concave down because placing any value in between these two numbers into  will provide a negative output.