All Calculus AB Resources
Example Questions
Example Question #1 : Calculate Higher Order Derivatives
True or False. The second derivative of a function is positive at . The function is concave down at .
False
True
False
The sign of the second derivative tells us what the concavity of the original function is. If the second derivative is positive, then the function is concave up. If the second derivative is negative, the function is concave down. So if the second derivative is positive, then the function would be concave up.
Example Question #6 : Calculate Higher Order Derivatives
Find the fourth order derivative of the function .
We will solve for this by finding the higher order derivatives up until we reach the fourth order derivative.
Example Question #1 : Calculate Higher Order Derivatives
Evaluate the third order derivative for at .
The third order derivative does not exist
The third order derivative does not exist
We begin by finding the third order derivative.
And since the third order derivative is it does not exist. So this solution does not exist.
Example Question #231 : Calculus Ab
On a closed interval, the function is decreasing. What can we say about and on these intervals?
is negative
Two or more of the other answers are correct.
is decreasing
is decreasing
is negative
is negative
If is decreasing, then its derivative is negative. The derivative of is , so this is telling us that is negative.
For to be decreasing, would have to be negative, which we don't know.
being negative has nothing to do with its slope.
For to be decreasing, its derivative would need to be negative, or, alternatively would have to be concave down, which we don't know.
Thus, the only correct answer is that is negative.
Example Question #232 : Calculus Ab
On what intervals is the function both concave up and decreasing?
The question is asking when the derivative is negative and the second derivative is positive. First, taking the derivative, we get
Solving for the zero's, we see hits zero at and . Constructing an interval test,
we want to know the sign's in each of these intervals. Thus, we pick a value in each of the intervals and plug it into the derivative to see if it's negative or positive. We've chosen and to be our three values.
Thus, we can see that the derivative is only negative on the interval .
Repeating the process for the second derivative,
The reader can verify that this equation hits at . Thus, the intervals to test for the second derivative are
. Plugging in and , we can see that the first interval is negative and the second is positive.
Because we want the interval where the second derivative is positive and the first derivative is negative, we need to take the intersection or overlap of the two intervals we got:
If this step is confusing, try drawing it out on a number line -- the first interval is from to , the second from to infinity. They only overlap on the smaller interval of to .
Thus, our final answer is
Example Question #41 : Differentiating Functions
If
and ,
then find .
We see the answer is when we use the product rule.
Example Question #1 : Calculate Higher Order Derivatives
If
and ,
then find .
We see the answer is when we use the product rule.
Example Question #2 : Calculate Higher Order Derivatives
On a closed interval, the function is decreasing. What can we say about and on these intervals?
is negative
is decreasing
is decreasing
is negative
Two or more of the other answers are correct.
is negative
If is decreasing, then its derivative is negative. The derivative of is , so this is telling us that is negative.
For to be decreasing, would have to be negative, which we don't know.
being negative has nothing to do with its slope.
For to be decreasing, its derivative would need to be negative, or, alternatively would have to be concave down, which we don't know.
Thus, the only correct answer is that is negative.
Example Question #3 : Calculate Higher Order Derivatives
On what intervals is the function both concave up and decreasing?
The question is asking when the derivative is negative and the second derivative is positive. First, taking the derivative, we get
Solving for the zero's, we see hits zero at and . Constructing an interval test,
, we want to know the sign's in each of these intervals. Thus, we pick a value in each of the intervals and plug it into the derivative to see if it's negative or positive. We've chosen -5, 0, and 1 to be our three values.
Thus, we can see that the derivative is only negative on the interval .
Repeating the process for the second derivative,
The reader can verify that this equation hits 0 at -4/3. Thus, the intervals to test for the second derivative are
. Plugging in -2 and 0, we can see that the first interval is negative and the second is positive.
Because we want the interval where the second derivative is positive and the first derivative is negative, we need to take the intersection or overlap of the two intervals we got:
If this step is confusing, try drawing it out on a number line -- the first interval is from -3 to 1/3, the second from -4/3 to infinity. They only overlap on the smaller interval of -4/3 to 1/3.
Thus, our final answer is .
Example Question #1 : Calculate Higher Order Derivatives
What does the second derivative of a function tell us?
The sign of the second derivative tells us if the function is concave up or down
The sign of the second derivative tells us if the first derivative is increasing or decreasing
All of the above are true
The sign of the second derivative tells us if the function is increasing or decreasing
All of the above are true
The second derivative is essentially a derivative of the first derivative. Information we get from the first derivative is, if the first derivative is increasing, then the function is also increasing and concave up. If the second derivative has a positive sign, then the first derivative is increasing and the function is concave up. Similarly if the function is concave down, the first derivative is decreasing, and the second derivative is negative.
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