### All AP Calculus AB Resources

## Example Questions

### Example Question #17 : Applications Of Derivatives

Find the coordinates of all local extrema for , and specify whether each is a local maximum or local minimum.

**Possible Answers:**

is a local maximum.

is a local minimum.

No local extrema

is a local maximum.

is a local minimum.

is a local maximum.

is a local minimum.

**Correct answer:**

is a local maximum.

is a local minimum.

To find the coordinates of the local extrema of a function, we need to find the critical points of its first derivative.

Since is a polynomial, we can find its derivative term by term. The first 3 terms can be differentiated using the power rule, , and the constant multiple rule, .

The last term is a constant, and its derivative is zero.

Applying these rules, we find the first derivative:

Now we need to find the critical points. To do this, we set the first derivative equal to zero and solve. Factoring is the best method in this problem.

Now that we have the critical points, we need to determine for each one, whether it is a maximum, minimum, or neither. We use the first derivative line test to determine this.

For , we will test the interval before it, , and the interval after it, , and find whether they are increasing or decreasing.

For the interval , we will test and find whether is positive or negative.

Since is positive, the original function is increasing before the critical point, .

Now we will test the interval after .

For the interval, , we will test and find whether is positive or negative.

Since is negative, the original function is decreasing in the interval following .

Since the function is increasing before , and decreasing afterward, we can conclude that a maximum occurs at .

Now we find the value of this maximum, .

Thus is a local maximum.

Now we will determine whether a maximum or minimum occurs at .

We know that is decreasing before , but we still need to determine what happens afterward.

For the interval , we will test , and find whether is positive or negative.

Since is positive, the original function is increasing following .

Since is decreasing before, and increasing after , we can conclude that a minimum occurs at .

Now we need to find the value of this minimum, .

Thus is a local minimum.

So our answer is:

is a local maximum.

is a local minimum.

### Example Question #18 : Applications Of Derivatives

Find the intervals of concavity for the function

**Possible Answers:**

Concave down:

Concave up:

Concave down:

Concave up:

Concave down:

Concave up:

Concave down:

Concave up:

Concave down:

Concave up:

**Correct answer:**

Concave down:

Concave up:

Concavity refers to the "curving" of the function. While the first derivative describes when the function is increasing or decreasing (instantaneous rate of change of ), the second derivative describes concavity, the instantaneous rate of change of .

The first derivative is like velocity, (moving forward or backward), while the second derivative is like acceleration (speeding up or slowing down).

Since we need to find the intervals of concavity, we will find the second derivative and work with it.

First we must find the first derivative using the power rule and constant multiple rule for each term of .

This gives:

Now to find the second derivative, we take the derivative of .

The same derivative rules apply:

Now that we have the second derivative, , we must find its critical points. We do this by setting , and solving.

The best method for this case is factoring.

The greatest common factor is ,

Inside the parentheses is a quadratic expression that can be factored like so:

Setting each factor equal to zero we find the following:

Now we know the critical points for the second derivative. To find the intervals of concavity, we test a point in each interval around these critical points and find whether the second derivative is positive or negative in that interval.

For the interval , we can test .

Using the factored form of the second derivative is easier than using the polynomial form, since the arithmetic involves fewer large numbers.

Since is negative, the original function is Concave Down in the interval .

Now for the interval , we can test .

We will combine the fractions inside the parentheses by getting the common denominator.

Multiplying the fractions gives:

Since is positive, the original function is Concave Up on the interval .

For the interval , we can test .

Since is negative, the original function is Concave Down on the interval .

Finally, for the interval , we can test .

Since is positive, the original function is Concave Up on the interval .

Summarizing the results, the intervals of concavity are:

Concave Down:

Concave Up:

### Example Question #1 : Analysis Of Curves, Including The Notions Of Monotonicity And Concavity

At the point where , is increasing or decreasing, and is it concave up or down?

**Possible Answers:**

There is no concavity at that point.

Increasing, concave down

Decreasing, concave down

Increasing, concave up

Decreasing, concave up

**Correct answer:**

Increasing, concave up

To find if the equation is increasing or decreasing, we need to look at the first derivative. If our result is positive at , then the function is increasing. If it is negative, then the function is decreasing.

To find the first derivative for this problem, we can use the power rule. The power rule states that we lower the exponent of each of the variables by one and multiply by that original exponent.

Remember that anything to the zero power is one.

Plug in our given value.

Is it positive? Yes. Then it is increasing.

To find the concavity, we need to look at the second derivative. If it is positive, then the function is concave up. If it is negative, then the function is concave down.

Repeat the process we used for the first derivative, but use as our expression.

For this problem, we're going to say that since, as stated before, anything to the zero power is one.

Notice that as anything times zero is zero.

As you can see, there is no place for a variable here. It doesn't matter what point we look at, the answer will always be positive. Therefore this graph is always concave up.

This means that at our given point, the graph is increasing and concave up.

### Example Question #20 : Applications Of Derivatives

Consider the function:

On what intervals is increasing? Consider all real numbers.

**Possible Answers:**

**Correct answer:**

To answer this question, one first needs to find and then find the critical points of the function (i.e. where . Finally, one would need to determine the sign of for the intervals between the critical points.

For the given function:

.

Therefore, when and . So, the intervals to consider are:

To determine the sign of , pick any number for the given interval and evaluate at that number.

Therefore, is increasing on the intervals and since is greater than zero on these intervals.

### Example Question #2 : Analysis Of Curves, Including The Notions Of Monotonicity And Concavity

A function, , is concave up on the intervals and with and .

Which of the following must be true?

**Possible Answers:**

Two or more of the other answers.

**Correct answer:**

On the domain , we know that the derivative begins positive, and because the concavity is positive, we know the derivative is increasing. Thus, the derivative stays positive for this entire interval, and the function increases from 2 to 4. Thus, must be greater than .

In the case of the interval , we know that the derivative is increasing, but it starts out negative. Thus, perhaps the derivative only increased from -1 to -0.5 in this interval, and the function would have decreased the entire time. In this case, would be less than , so we can't really say anything about these values.

For the remaining two, there's not any clear way to relate the functions at and . While we know needs to be bigger than 1, we don't know by how much. Similarly, while we know needs to be bigger than -1, we don't know by how much. Thus, it's completely possible that and .

As for and , we know even less. In between the two intervals, our function could have shot up a million, or shot down by the same amount. Thus, there's no safe comparison we can make between these two values.

### Example Question #3 : Analysis Of Curves, Including The Notions Of Monotonicity And Concavity

Find the intervals on which is increasing.

**Possible Answers:**

**Correct answer:**

To find the intervals where the function is increasing, we need to find the points at which its slope changes from positive to negative and vice versa. The first derivative, which is the slope at any point, will help us.

First, we find the derivative of , using the power rule for each term. Recall that the power rule says

Also, the constant multiple rule will apply to the coefficients of each term. The constant multiple rule simply says that any constant factor of a term will "carry" to the derivative of that term. For example:

Lastly, the derivative of a constant is zero. This will result in the last term, , dropping off as we take the derivative.

Applying these rules, we find the derivative

Where the derivative is positive (blue line in graph), the tangent line to the original function is angled up. Where the derivative is negative (red line in graph), the slope of the tangent line is angled down.

The points where the tangent line's slope transitions from negative to positive or vice versa, are called the critical points. At these points, the tangent line becomes a horizontal line with a slope of zero (green line in graph). In other words, the "critical points" occur when the derivative is zero. These points will be the endpoints of our intervals of increasing and decreasing. To find the critical points, we will set the derivative equal to zero and solve for x. In this problem factoring is the best method:

Now that we have found the critical points, we need to know whether the original function is increasing or decreasing in the intervals between them. We will do so by testing a point in each interval and determining whether the derivative is positive or negative at that point. This is called the first derivative line test.

For the interval , we will test .

(Note: I will use the factored form of the derivative, but we could also use the polynomial version. Both will give the same result)

Since the derivative is negative at this point, we can conclude that the derivative is negative for the whole interval. Thus, the original function is decreasing on .

For the interval , we will test .

Since the derivative is positive at this point, we can again conclude that the derivative is positive for the whole interval. Thus, the original function is increasing on .

For the interval , we will test .

Since the derivative is negative at this point, we can again conclude that the derivative is negative for the whole interval. Thus, the original function is decreasing on .

Lastly, for the interval , we will test .

Since the derivative is positive at this point, we know that the derivative is positive for the whole interval. Thus the original function is increasing on .

From these 4 results, we now know the answer. The function is increasing on the intervals .

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