How to find concave down intervals by graphing functions - AP Calculus AB

To find when a function is concave, you must first take the 2nd derivative, then set it equal to 0, and then find between which zero values the function is negative.

First, find the 2nd derivative:

Set equal to 0 and solve:

Now test values on all sides of these to find when the function is negative, and therefore decreasing. I will test the values of -3 and 0.

Since the value that is negative is when x=-3, the interval is decreasing on the interval that includes 0. Therefore, our answer is:

To find when a function is concave, you must first take the 2nd derivative, then set it equal to 0, and then find between which zero values the function is negative.

First, find the 2nd derivative:

Set equal to 0 and solve:

Now test values on all sides of these to find when the function is negative, and therefore decreasing. I will test the values of 0 and 6.

Since the value that is negative is when x=0, the interval is decreasing on the interval that includes 0. Therefore, our answer is:

To test concavity, we must first find the second derivative of f(x)

This function is concave down anywhere that f''(x)<0, so...

So,

for all

So on the interval -5,-4 f(x) is concave down because f''(x) is negative.

To test concavity, we need to perform the second derivative test. Bascially, assuming that our function is twice-differentiable, anywhere that h"(t) is positive will be concave up, and anywhere h"(t) is negative is concave down. Begin as follows:

Next, we need to evaluate h"(t) on the interval \[5,7\]

So, our h"(t) is positive on the interval, and therefore h(t) is concave up.

The intervals where a function is concave up or down is found by taking second derivative of the function. Use the power rule which states:

Now, set equal to to find the point(s) of infleciton.

In this case, .

To find the concave up region, find where is positive. This will either be to the left of or to the right of . To find out which, plug in a test point in each of those regions.

If we plug in we get , which is negative, so that cannot be concave up.

If we plug in , we get , which is positive, so we know that the region will be concave up.

To find which interval is concave down, find the second derivative of the function.

Now, find which values in the interval specified make . In this case, and .

Now to find which interval is concave down choose any value in each of the regions

, and

and plug in those values into to see which will give a negative answer, meaning concave down, or a positive answer, meaning concave up.

A test value of gives us a of . This value falls in the range, meaning that interval is concave down.

The derivative of is

The derivative of this is

This is the second derivative.

A function is concave down if its second derivative is less than 0.

whenever

This is true when:

To find the concavity of a graph, the double derivative of the graph equation has to be taken. To take the derivative of this equation, we must use the power rule,

.

We also must remember that the derivative of an constant is 0.

After taking the first derivative of the equation using the power rule, we obtain

.

The double derivative of the equation we are given comes out to

.

Setting the equation equal to zero, we find that . This point is our inflection point, where the graph changes concavity. In order to find what concavity it is changing from and to, you plug in numbers on either side of the inflection point. if the result is negative, the graph is concave down and if it is positive the graph is concave up. Plugging in 2 and 3 into the second derivative equation, we find that the graph is concave up from and concave down from .

To find the invervals where a function is concave down, you must find the intervals on which the second derivative of the function is negative. To find the intervals, first find the points at which the second derivative is equal to zero. The first derivative of the function is equal to

.

The second derivative of the function is equal to

.

Both derivatives were found using the power rule

.

Solving for x, . The intervals, therefore, that we analyze are and .

On the first interval, the second derivative is negative, which means the function is concave down. On the second interval, the second derivative is positive, which means the function is concave up. (Plug in values on the intervals into the second derivative and see if they are positive or negative.)

Thus, the first interval is the answer.

Is the following function concave up or concave down on the interval \[2,4\]?

A function is concave up if its second derivative is positive, and vice-versa. so, we want to find h''(t).

Next, plug in the endpoints of our interval to find the sign of h''(t).

Since our second derivative is negative on the interval, we can say that h(t) is concave down on the interval.

Points of inflection occur where there second derivative of a function are equal to zero. Taking the first and second derivative of the function, we find:

To find the points of inflection, we find the values of x that satisfy the condition

.

Which occurs at

Within the defined interval \[-5, 5\], there are three values: . These points are represented on the figure below as red dots.

To find the concavity of a function, we must take the second derivative of the function and set it equal to zero.

Solving for , the point of inflection occurs at :

We must find where is negative, which will either be to the left or the right of . Test points can be used to determine this.

Because the region less than is negative, we know that the region from is concave down.

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.