### All Calculus 1 Resources

## Example Questions

### Example Question #6 : Derivatives

Find the inflection point(s) of .

**Possible Answers:**

**Correct answer:**

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 #74 : Graphing Functions

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

**Possible Answers:**

Not enough information

**Correct answer:**

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 #78 : Graphing Functions

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

**Possible Answers:**

and

and

There are no points of inflection.

**Correct answer:**

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 #79 : Graphing Functions

What are the coordinates of the points of inflection for the graph

**Possible Answers:**

There are no points of inflection on this graph.

**Correct answer:**

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 .

### Example Question #75 : Graphing Functions

Find all the points of inflection of

.

**Possible Answers:**

There are no inflection points.

**Correct answer:**

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 #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:**

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 #2 : How To Graph Functions Of Points Of Inflection

Find all the points of infection of

.

**Possible Answers:**

There are no points of inflection.

**Correct answer:**

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 #3 : How To Graph Functions Of 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.

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 #82 : Graphing Functions

Find the points of inflection of the following function:

**Possible Answers:**

**Correct answer:**

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 #83 : Graphing Functions

Find all the points of inflection of

.

**Possible Answers:**

**Correct answer:**

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.