Functions - AP Calculus AB

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 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 .

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.

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.

Is the following function concave up or down on the interval ?

Recall that the test for cancavity is similar to the one for increasing/decreasing, but with a key difference. To test concavity, we must use the second derivative instead of the first.

Begin by putting f(x) in standard form and then find its derivative

Now that we have our second derivative, we need to try both our endpoints to see what kind of an answer we get:

So, both our endpoints yield negative numbers. If you think critically, you should be able to see that any number between 3 and 6 will also yield a negative number.

This means that f(x) is concave down on the given interval, because all possible values of its second derivative on the interval are negative.

To find the intervals with the same concavity, we need to find the critical points using the second derivative test, then see what the concavity is in the intervals using the second derivative.

; set equal to 0 and solve for , giving as the only critical point.

Choose an -value either side of the critical point, and test the concavity. For example:

, so the graph is concave down to the left of the critical point.

, so the graph is concave up to the right of the critical point.

Therefore the function is concave down on the interval .

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

First, we must find the second derivative:

The derivatives were found using the following rule:

Now, we set the second derivative equal to zero and find the x values at which this is true:

Next, we make our intervals on which we determine if the second derivative is positive or negative, with 0 as the upper and lower bound for the intervals:

Note that at 0 the second derivative is neither positive nor negative.

On the first interval, the second derivative is always negative, while on the second interval, the second derivative is always positive (simply plug in any point on the interval into the second derivative and check the sign). Thus, on the first interval - - the function is concave down.

To solve, simply differentiate twice, using the power rule, and then find where the second derivative is negative.

Power rule:

Thus,

Since the second derivative is always negative, our function is concave down everywhere.

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

First, we must find the second derivative:

The derivatives were found using the following rule:

Now, we must find the value at which the second derivative is equal to zero.

We will now use this as the upper and lower limit of our intervals on which we evaluate the sign of the second derivative:

On the first interval, the second derivative is negative, while on the second interval, the second derivative is positive. Thus, our answer is .

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

First, we must find the first and second derivatives:

The derivatives were found using the following rule:

Next, we must find the values at which the second derivative is equal to zero:

Now, we can make the intervals:

Note that at the bounds of the intervals the second derivative is neither positive nor negative.

To determine the sign of the second derivative on the intervals, simply plug in any value on the interval into the second derivative function; on the first interval, the second derivative is positive, on the second it is negative, and on the third it is positive. Thus, the function is concave down on the interval .

Tell whether f(x) is concave up or concave down on the interval \[1,2\]

To find concave up and concave down, we need to find the second derivative of f(x).

Let's begin by finding f'(x)

Next find f ''(x)

Now, to test for concavity, plug in the endpoints of the interval:

So, on this interval, f"(x) will always be negative. This means that our function is concave down on this interval.

Is the function b(t) concave up, concave down, or neither when t is equal to -3?

To test for concavity, we need to find the sign of the function's second derivative at the given time.

Begin by recalling that the derivative of a polynomial is found by multiplying each term by its exponent, then decreasing the exponent by 1.

Doing this gets us the following:

Almost there, but we need b"(-3)

b"(-3) is negative, therefore our function is concave down when t=-3

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

First, we must find the second derivative of the function:

The derivatives were found using the following rules:

, ,

Note that for the first rule, the chain rule, as it applies to the natural log, regardless of the constant in front of x, the derivative of the natural log will always be the same.

Next, we must find the values at which the second derivative is equal to zero. This never occurs, so there is no place at which the second derivative is equal to zero. Furthermore, the second derivative is always positive, so the function is never concave down.

To find concavity, you need to find the second derivative. In order to find that, you must first determine the first derivative. When taking the derivative, multiply the exponent by the coefficent in front of the x and then subtract one from the exponent. The first derivative is . Then, find the second derivative is . Set that equal to 0 to get your critical point: Then, test values on either side of 0 by plugging into the second derivative. To the left of 0, the value is negative. To the right, the value is positive. Therfore, it is concave up from .