Inflection Points - AP Calculus BC

Inflection points can only occur when the second derivative is zero or undefined. Here we have

.

Therefore possible inflection points occur at and . However, to have an inflection point we must check that the sign of the second derivative is different on each side of the point. Here we have

.

Hence, both are inflection points

Possible inflection points occur when . This occurs at three values, . However, to be an inflection point the sign of must be different on either side of the critical value. Hence, only are critical points.

A point of inflection is found where the graph (or image) of a function changes concavity. To find this algebraically, we want to find where the second derivative of the function changes sign, from negative to positive, or vice-versa. So, we find the second derivative of the given function

The first derivative using the power rule

is,

and the seconds derivative is

We then find where this second derivative equals . when .

We then look to see if the second derivative changes signs at this point. Both graphically and algebraically, we can see that the function does indeed change sign at, and only at, , so this is our inflection point.

Infelction points are the points of a graph where the concavity of the graph changes. The inflection points of a graph are found by taking the double derivative of the graph equation, setting it equal to zero, then solving for .

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 graph equation using the power rule, the equation becomes

.

In this problem the double derivative of the graph equation comes out to , factoring this equation out it becomes .

Solving for when the equation is set equal to zero, the inflection points are located at .

Even though the first derivative () is at , there is no max or min because the function is increasing on both sides (derivative is positive on both sides). However, the function is changing its concavity (from negative to positive) at . We can tell because the second derivative () is also at , and it's going from negative to positive. Hole and asymptote are irrelevant. When the second derivative changes signs around a specific point we call this an inflection point. An inflection point describes a point that changes the concavity of a function.

In order to find the points of inflection, we need to find using the power rule .

Now to find the points of inflection, we need to set .

.

Now we can use the quadratic equation.

Recall that the quadratic equation is

,

where a,b,c refer to the coefficients of the equation .

In this case, a=12, b=0, c=-4.

Thus the possible points of infection are

.

Now to check if or which are inflection points we need to plug in a value higher and lower than each point. If there is a sign change then the point is an inflection point.

To check lets plug in .

Therefore is an inflection point.

Now lets check with .

Therefore is also an inflection point.

In order to find the points of inflection, we need to find using the power rule .

Now lets factor .

Now to find the points of inflection, we need to set .

.

From this equation, we already know one of the point of inflection, .

To figure out the rest of the points of inflection we can use the quadratic equation.

Recall that the quadratic equation is

, where a,b,c refer to the coefficients of the equation

.

In this case, a=20, b=0, c=-18.

Thus the other 2 points of infection are

To verify that they are all inflection points we need to plug in values higher and lower than each value and see if the sign changes.

Lets plug in

Since there is a sign change at each point, all are points of inflection.

Which of the following is a point of inflection of f(x) on the interval ?

To find points of inflection, we need to know where the second derivative of the function is equal to zero. So, find the second derivative:

So, where on the given interval does ?

Well, we know from our unit circle that ,

So we would have a point of inflection at , but we still need to find the y-coordinate of our POI. find this by finding

So our POI is:

Since the function is increasing at , , and since is below the x-axis, .

Furthermore, there exists an inflection point at , where the concavity of function changes.

Thus at , .

Therefore, the correct answer is .

For a graph to have an inflection point, the second derivative must be equal to zero. We also want the concavity to change at that point.

, for all real numbers, but this is a line and has no concavity associated with it, so not this one.

, there are no real values of for which this equals zero, so no inflection points.

, same story here.

, so no inflection points here.

This leaves us with

, whose derivatives are a bit more difficult to take.

, so by the chain rule we get

So, when . So

. This is our correct answer.