Corresponding characteristics of graphs of ƒ and ƒ' - AP Calculus AB
Card 1 of 55
A gun sends a bullet straight up with a launch velocity of 220 ft/s. It reaches a height of $s=220t-16t^2$ after 
seconds. What's the acceleration in 
of the block after it has been ejected?
A gun sends a bullet straight up with a launch velocity of 220 ft/s. It reaches a height of $s=220t-16t^2$ after
Tap to reveal answer
Since $a=\frac{ds^2$$}{dt^2$}$, by differentiating the position function twice, we see that acceleration is constant and 
. Acceleration, in this case, is gravity, which makes sense that it should be a constant value!
Since $a=\frac{ds^2$$}{dt^2$}$, by differentiating the position function twice, we see that acceleration is constant and
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A jogger leaves City 
at 
. His subsequent position, in feet, is given by the function:
,
where 
is the time in minutes.
Find the acceleration of the jogger at 
minutes.
A jogger leaves City
where
Find the acceleration of the jogger at
Tap to reveal answer
The accelaration is given by the second derivative of the position function:
For the given position function:
,
,
.
Therefore, the acceleration at 
minutes is 
. Again, note the units must be in 
.
The accelaration is given by the second derivative of the position function:
For the given position function:
Therefore, the acceleration at
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The speed of a car traveling on the highway is given by the following function of time:
Consider a second function:
What can we conclude about this second function?
The speed of a car traveling on the highway is given by the following function of time:
Consider a second function:
What can we conclude about this second function?
Tap to reveal answer
Notice that the function 
is simply the derivative of 
with respect to time. To see this, simply use the power rule on each of the two terms.
Therefore, 
is the rate at which the car's speed changes, a quantity called acceleration.
Notice that the function
Therefore,
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The function 
is a continuous, twice-differentiable functuon defined for all real numbers.
If the following are true:
Which function could be 
?
The function
If the following are true:
Which function could be
Tap to reveal answer
To answer this problem we must first interpret our given conditions:
Implies the function is strictly increasing.
Implies the function is strictly concave down.
We note the only function given which fufills both of these conditions is 
.
To answer this problem we must first interpret our given conditions:
We note the only function given which fufills both of these conditions is
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Find the critical numbers of the function,
Find the critical numbers of the function,
Tap to reveal answer
1) Recall the definition of a critical point:
The critical points of a function 
are defined as points 
, such that 
is in the domain of , and at which the derivative 
is either zero or does not exist. The number 
is called a critical number of 
.
2) Differentiate 
,
3) Set to zero and solve for 
,
The critical numbers are,
We can also observe that the derivative does not exist at 
, since the 
term would be come infinite. However, 
is not a critical number because the original function 
is not defined at 
. The original function is infinite at 
, and therefore 
is a vertical asymptote of 
as can be seen in its' graph,
Further Discussion
In this problem we were asked to obtain the critical numbers. If were were asked to find the critical points, we would simply evaluate the function 
at the critical numbers to find the corresponding function values and then write them as a set of ordered pairs,
1) Recall the definition of a critical point:
The critical points of a function
2) Differentiate
3) Set to zero and solve for
The critical numbers are,
We can also observe that the derivative does not exist at
Further Discussion
In this problem we were asked to obtain the critical numbers. If were were asked to find the critical points, we would simply evaluate the function
← Didn't Know|Knew It →
A gun sends a bullet straight up with a launch velocity of 220 ft/s. It reaches a height of $s=220t-16t^2$ after seconds. What's the acceleration in of the block after it has been ejected?
A gun sends a bullet straight up with a launch velocity of 220 ft/s. It reaches a height of $s=220t-16t^2$ after
Tap to reveal answer
Since $a=\frac{ds^2$$}{dt^2$}$, by differentiating the position function twice, we see that acceleration is constant and . Acceleration, in this case, is gravity, which makes sense that it should be a constant value!
Since $a=\frac{ds^2$$}{dt^2$}$, by differentiating the position function twice, we see that acceleration is constant and
← Didn't Know|Knew It →
A jogger leaves City at . His subsequent position, in feet, is given by the function:
,
where is the time in minutes.
Find the acceleration of the jogger at minutes.
A jogger leaves City
where
Find the acceleration of the jogger at
Tap to reveal answer
The accelaration is given by the second derivative of the position function:
For the given position function:
,
,
.
Therefore, the acceleration at minutes is . Again, note the units must be in .
The accelaration is given by the second derivative of the position function:
For the given position function:
Therefore, the acceleration at
← Didn't Know|Knew It →
The speed of a car traveling on the highway is given by the following function of time:
Consider a second function:
What can we conclude about this second function?
The speed of a car traveling on the highway is given by the following function of time:
Consider a second function:
What can we conclude about this second function?
Tap to reveal answer
Notice that the function is simply the derivative of with respect to time. To see this, simply use the power rule on each of the two terms.
Therefore, is the rate at which the car's speed changes, a quantity called acceleration.
Notice that the function
Therefore,
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The function is a continuous, twice-differentiable functuon defined for all real numbers.
If the following are true:
Which function could be ?
The function
If the following are true:
Which function could be
Tap to reveal answer
To answer this problem we must first interpret our given conditions:
- Implies the function is strictly increasing.
- Implies the function is strictly concave down.
We note the only function given which fufills both of these conditions is .
To answer this problem we must first interpret our given conditions:
We note the only function given which fufills both of these conditions is
← Didn't Know|Knew It →
Find the critical numbers of the function,
Find the critical numbers of the function,
Tap to reveal answer
1) Recall the definition of a critical point:
The critical points of a function are defined as points , such that is in the domain of , and at which the derivative is either zero or does not exist. The number is called a critical number of .
2) Differentiate ,
3) Set to zero and solve for ,
The critical numbers are,
We can also observe that the derivative does not exist at , since the term would be come infinite. However, is not a critical number because the original function is not defined at . The original function is infinite at , and therefore is a vertical asymptote of as can be seen in its' graph,
Further Discussion
In this problem we were asked to obtain the critical numbers. If were were asked to find the critical points, we would simply evaluate the function at the critical numbers to find the corresponding function values and then write them as a set of ordered pairs,
1) Recall the definition of a critical point:
The critical points of a function
2) Differentiate
3) Set to zero and solve for
The critical numbers are,
We can also observe that the derivative does not exist at
Further Discussion
In this problem we were asked to obtain the critical numbers. If were were asked to find the critical points, we would simply evaluate the function
← Didn't Know|Knew It →
Find the critical numbers of the function,
Find the critical numbers of the function,
Tap to reveal answer
1) Recall the definition of a critical point:
The critical points of a function are defined as points , such that is in the domain of , and at which the derivative is either zero or does not exist. The number is called a critical number of .
2) Differentiate ,
3) Set to zero and solve for ,
The critical numbers are,
We can also observe that the derivative does not exist at , since the term would be come infinite. However, is not a critical number because the original function is not defined at . The original function is infinite at , and therefore is a vertical asymptote of as can be seen in its' graph,
Further Discussion
In this problem we were asked to obtain the critical numbers. If were were asked to find the critical points, we would simply evaluate the function at the critical numbers to find the corresponding function values and then write them as a set of ordered pairs,
1) Recall the definition of a critical point:
The critical points of a function
2) Differentiate
3) Set to zero and solve for
The critical numbers are,
We can also observe that the derivative does not exist at
Further Discussion
In this problem we were asked to obtain the critical numbers. If were were asked to find the critical points, we would simply evaluate the function
← Didn't Know|Knew It →
A gun sends a bullet straight up with a launch velocity of 220 ft/s. It reaches a height of $s=220t-16t^2$ after seconds. What's the acceleration in of the block after it has been ejected?
A gun sends a bullet straight up with a launch velocity of 220 ft/s. It reaches a height of $s=220t-16t^2$ after
Tap to reveal answer
Since $a=\frac{ds^2$$}{dt^2$}$, by differentiating the position function twice, we see that acceleration is constant and . Acceleration, in this case, is gravity, which makes sense that it should be a constant value!
Since $a=\frac{ds^2$$}{dt^2$}$, by differentiating the position function twice, we see that acceleration is constant and
← Didn't Know|Knew It →
A jogger leaves City at . His subsequent position, in feet, is given by the function:
,
where is the time in minutes.
Find the acceleration of the jogger at minutes.
A jogger leaves City
where
Find the acceleration of the jogger at
Tap to reveal answer
The accelaration is given by the second derivative of the position function:
For the given position function:
,
,
.
Therefore, the acceleration at minutes is . Again, note the units must be in .
The accelaration is given by the second derivative of the position function:
For the given position function:
Therefore, the acceleration at
← Didn't Know|Knew It →
The speed of a car traveling on the highway is given by the following function of time:
Consider a second function:
What can we conclude about this second function?
The speed of a car traveling on the highway is given by the following function of time:
Consider a second function:
What can we conclude about this second function?
Tap to reveal answer
Notice that the function is simply the derivative of with respect to time. To see this, simply use the power rule on each of the two terms.
Therefore, is the rate at which the car's speed changes, a quantity called acceleration.
Notice that the function
Therefore,
← Didn't Know|Knew It →
The function is a continuous, twice-differentiable functuon defined for all real numbers.
If the following are true:
Which function could be ?
The function
If the following are true:
Which function could be
Tap to reveal answer
To answer this problem we must first interpret our given conditions:
- Implies the function is strictly increasing.
- Implies the function is strictly concave down.
We note the only function given which fufills both of these conditions is .
To answer this problem we must first interpret our given conditions:
We note the only function given which fufills both of these conditions is
← Didn't Know|Knew It →
A gun sends a bullet straight up with a launch velocity of 220 ft/s. It reaches a height of $s=220t-16t^2$ after seconds. What's the acceleration in of the block after it has been ejected?
A gun sends a bullet straight up with a launch velocity of 220 ft/s. It reaches a height of $s=220t-16t^2$ after
Tap to reveal answer
Since $a=\frac{ds^2$$}{dt^2$}$, by differentiating the position function twice, we see that acceleration is constant and . Acceleration, in this case, is gravity, which makes sense that it should be a constant value!
Since $a=\frac{ds^2$$}{dt^2$}$, by differentiating the position function twice, we see that acceleration is constant and
← Didn't Know|Knew It →
A jogger leaves City at . His subsequent position, in feet, is given by the function:
,
where is the time in minutes.
Find the acceleration of the jogger at minutes.
A jogger leaves City
where
Find the acceleration of the jogger at
Tap to reveal answer
The accelaration is given by the second derivative of the position function:
For the given position function:
,
,
.
Therefore, the acceleration at minutes is . Again, note the units must be in .
The accelaration is given by the second derivative of the position function:
For the given position function:
Therefore, the acceleration at
← Didn't Know|Knew It →
The speed of a car traveling on the highway is given by the following function of time:
Consider a second function:
What can we conclude about this second function?
The speed of a car traveling on the highway is given by the following function of time:
Consider a second function:
What can we conclude about this second function?
Tap to reveal answer
Notice that the function is simply the derivative of with respect to time. To see this, simply use the power rule on each of the two terms.
Therefore, is the rate at which the car's speed changes, a quantity called acceleration.
Notice that the function
Therefore,
← Didn't Know|Knew It →
The function is a continuous, twice-differentiable functuon defined for all real numbers.
If the following are true:
Which function could be ?
The function
If the following are true:
Which function could be
Tap to reveal answer
To answer this problem we must first interpret our given conditions:
- Implies the function is strictly increasing.
- Implies the function is strictly concave down.
We note the only function given which fufills both of these conditions is .
To answer this problem we must first interpret our given conditions:
We note the only function given which fufills both of these conditions is
← Didn't Know|Knew It →
Find the critical numbers of the function,
Find the critical numbers of the function,
Tap to reveal answer
1) Recall the definition of a critical point:
The critical points of a function are defined as points , such that is in the domain of , and at which the derivative is either zero or does not exist. The number is called a critical number of .
2) Differentiate ,
3) Set to zero and solve for ,
The critical numbers are,
We can also observe that the derivative does not exist at , since the term would be come infinite. However, is not a critical number because the original function is not defined at . The original function is infinite at , and therefore is a vertical asymptote of as can be seen in its' graph,
Further Discussion
In this problem we were asked to obtain the critical numbers. If were were asked to find the critical points, we would simply evaluate the function at the critical numbers to find the corresponding function values and then write them as a set of ordered pairs,
1) Recall the definition of a critical point:
The critical points of a function
2) Differentiate
3) Set to zero and solve for
The critical numbers are,
We can also observe that the derivative does not exist at
Further Discussion
In this problem we were asked to obtain the critical numbers. If were were asked to find the critical points, we would simply evaluate the function
← Didn't Know|Knew It →