Calculus 1 : Points of Inflection

Study concepts, example questions & explanations for Calculus 1

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Example Questions

Example Question #1 : Derivatives

Find the inflection point(s) of .

Possible Answers:

Correct answer:

Explanation:

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

Example Question #1 : Points Of Inflection

Below is the graph of . How many inflection points does  have?Graph1

Possible Answers:

Not enough information

Correct answer:

Explanation:

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.

Example Question #1 : Points Of Inflection

Find the point(s) of inflection for the function .

Possible Answers:

 and 

 and 

There are no points of inflection.

Correct answer:

Explanation:

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.

Example Question #1 : Points Of Inflection

Find all the points of inflection of

.

Possible Answers:

There are no inflection points.

Correct answer:

Explanation:

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

Now we set , and solve for .

To verify this is a true inflection point we need to plug in a value that is less than it and a value that is greater than it into the second derivative. If there is a sign change around the  point than it is a true inflection point.

Let 

Now let 

Since the sign changes from a positive to a negative around the point , we can conclude it is an inflection point.

Example Question #2 : Points Of Inflection

Find all the points of inflection of

Possible Answers:

There are no points of inflection.

Correct answer:

Explanation:

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. 

Example Question #1 : Points Of Inflection

Find all the points of infection of

.

Possible Answers:

There are no points of inflection.

Correct answer:

Explanation:

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.

Example Question #2 : Points Of Inflection

Find the points of inflection of

.

Possible Answers:

There are no points of inflection.

Correct answer:

There are no points of inflection.

Explanation:

In order to find the points of inflection, we need to find 

Now we set .

.

This last statement says that  will never be . Thus there are no points of inflection.

 

Example Question #1 : Points Of Inflection

Find the points of inflection of the following function:

Possible Answers:

Correct answer:

Explanation:

The points of inflection of a given function are the values at which the second derivative of the function are equal to zero.

The first derivative of the function is 

, and the derivative of this function (the second derivative of the original function), is 

.

Both derivatives were found using the power rule 

Solving .

 

To verify that this point is a true inflection point we need to plug in a value that is less than the point and one that is greater than the point into the second derivative. If there is a sign change between the two numbers than the point in question is an inflection point.

Lets plug in 

.

Now plug in 

.

Therefore,  is the only point of inflection of the function.

Example Question #1 : Points Of Inflection

Find all the points of inflection of

.

Possible Answers:

Correct answer:

Explanation:

In order to find all the points of inflection, we first find  using the power rule twice, .

Now we set .

.

Now we factor the left hand side.

From this, we see that there is one point of inflection at .

For the point of inflection, lets solve for x for the equation inside the parentheses. 

 

 

Example Question #1 : How To Graph Functions Of Points Of Inflection

 Find all the points of inflection of:

Possible Answers:

There are no points of inflection.

Correct answer:

Explanation:

 In order to find all the points of inflection, we first find  using the power rule twice .

Now we set .

Thus the points of inflection are   and 

 

 

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