### All Calculus 1 Resources

## Example Questions

### Example Question #5 : How To Graph Functions Of Area

Find the area bounded by the curve in the first quadrant.

**Possible Answers:**

**Correct answer:**

The curve is in quadrant one over the interval , which are the bounds of integration. To see this, note that the x-intercepts are 0 and 2 and the parabola opens downward.

The definite integral below is solved by taking the antiderivative of each term of the given polynomial function, evaluating this antiderivative at the bounds of integration, and subtracting the values.

For this particular integral use the rule, to solve.

### Example Question #6 : How To Graph Functions Of Area

Find the area under the curve between the following bounderies of the following function.

in between the boundaries of and

**Possible Answers:**

**Correct answer:**

We can find the area under the curve by taking the anti-derivative of the function and using the two boundaries as x values. The anti-derivative of is . If we use the two boundaries, we end up with our answer, .

### Example Question #1 : How To Graph Functions Of Area

Evalute the following Definite Integral:

**Possible Answers:**

**Correct answer:**

For this integral one may be tempted to directly integrate; however, this is no rule of integrals which allows us to do so.

We must apply a complex method of integration, here u-subtitution works best.

We notice that Cos(x) is the derivative of Sin(x) so it may be best to let .

Now that we have found a proper u and du, we may directly substiute into our original integral

However, our limits are in terms of x so we must substitute our u=Sin(x) back in before evaluating

### Example Question #1 : How To Describe Points By Graphing Functions

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?

**Possible Answers:**

**Correct answer:**

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 #2 : How To Describe Points By Graphing Functions

What is the critical point for ?

**Possible Answers:**

**Correct answer:**

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 #1 : Points Of Inflection

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 #2 : How To Graph Functions Of Points Of Inflection

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 #3 : How To Graph Functions Of Points Of Inflection

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

**Possible Answers:**

and

There are no points of inflection.

and

**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 #4 : How To Graph Functions Of Points Of Inflection

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 : 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.

Certified Tutor