# Calculus 1 : Points

## Example Questions

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### Example Question #1 : Describing Points

Suppose a point on the curve given above has the property that

Based solely on the graph above, which of the following is most likely the  value of the point in question?

Explanation:

If then the graph must be concave up at the point. Based on the picture, we know that the curve is concave up on  at best. The only value that falls on this interval is , which is . Since , this definitely falls on the interval given and we can be sure it is concave up based on the picture.

### Example Question #1 : Describing Points

What is the critical point for ?

Explanation:

To find the critical point, you must find the derivative first. To do that, multiply the exponent by the coefficient in front of the  and then subtract the exponent by . Therefore, the derivative is: . Then, to find the critical point, set the derivative equal to .

.

### Example Question #76 : Graphing Functions

Find the inflection point(s) of .

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 #77 : Graphing Functions

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

Not enough information

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 #71 : Graphing Functions

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

There are no points of inflection.

and

and

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 #1521 : Functions

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

There are no points of inflection on this graph.

Explanation:

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 #80 : Graphing Functions

Find all the points of inflection of

.

There are no inflection points.

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 #81 : Graphing Functions

Find all the points of inflection of

There are no points of inflection.

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 #82 : Graphing Functions

Find all the points of infection of

.

There are no points of inflection.

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 #1 : How To Graph Functions Of Points Of Inflection

Find the points of inflection of

.

There are no points of inflection.

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.

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