Corresponding characteristics of graphs of ƒ and ƒ' - AP Calculus AB

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Question

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 $m/s^{2}$ of the block after it has been ejected?

Answer

Since $a=\frac{ds^2}{dt^2}$ , by differentiating the position function twice, we see that acceleration is constant and $-32\ m/s^{2}$ . Acceleration, in this case, is gravity, which makes sense that it should be a constant value!

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Question

A jogger leaves City at . His subsequent position, in feet, is given by the function:

$x(t) = 5t^{2} - 3t + 2$ ,

where is the time in minutes.

Find the acceleration of the jogger at minutes.

Answer

The accelaration is given by the second derivative of the position function:

For the given position function:

$f(t) = 5t^{2} - 3t + 2$ ,

,

.

Therefore, the acceleration at minutes is $10\ ft/min^{2}$ . Again, note the units must be in $ft/min^{2}$ .

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Question

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?

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.

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Question

Find the critical numbers of the function,

$f(x)=\frac{1+x^2}{x}$

Answer

$f(x)=\frac{1+x^2}{x}$

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 ,

$f'(x)=\frac{d}{dx}\left(\frac{1+x^2}{x} \right )=\frac{d}{dx}\left(\frac{1}{x}+x \right )=-\frac{1}{x^2}+1$

3) Set to zero and solve for ,

$f'(x)=-\frac{1}{x^2}+1= 0$

$\frac{1}{x^2}=1$

The critical numbers are,

$x =\left { -1,1\right }$

We can also observe that the derivative does not exist at , since the $-\frac{1}{x^2}$ 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,

$\left {(-1,-2), (1,2) \right }$

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Question

The function is a continuous, twice-differentiable functuon defined for all real numbers.

If the following are true:

Which function could be ?

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 .

Compare your answer with the correct one above

Question

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 $m/s^{2}$ of the block after it has been ejected?

Answer

Since $a=\frac{ds^2}{dt^2}$ , by differentiating the position function twice, we see that acceleration is constant and $-32\ m/s^{2}$ . Acceleration, in this case, is gravity, which makes sense that it should be a constant value!

Compare your answer with the correct one above

Question

A jogger leaves City at . His subsequent position, in feet, is given by the function:

$x(t) = 5t^{2} - 3t + 2$ ,

where is the time in minutes.

Find the acceleration of the jogger at minutes.

Answer

The accelaration is given by the second derivative of the position function:

For the given position function:

$f(t) = 5t^{2} - 3t + 2$ ,

,

.

Therefore, the acceleration at minutes is $10\ ft/min^{2}$ . Again, note the units must be in $ft/min^{2}$ .

Compare your answer with the correct one above

Question

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?

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.

Compare your answer with the correct one above

Question

Find the critical numbers of the function,

$f(x)=\frac{1+x^2}{x}$

Answer

$f(x)=\frac{1+x^2}{x}$

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 ,

$f'(x)=\frac{d}{dx}\left(\frac{1+x^2}{x} \right )=\frac{d}{dx}\left(\frac{1}{x}+x \right )=-\frac{1}{x^2}+1$

3) Set to zero and solve for ,

$f'(x)=-\frac{1}{x^2}+1= 0$

$\frac{1}{x^2}=1$

The critical numbers are,

$x =\left { -1,1\right }$

We can also observe that the derivative does not exist at , since the $-\frac{1}{x^2}$ 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,

$\left {(-1,-2), (1,2) \right }$

Compare your answer with the correct one above

Question

The function is a continuous, twice-differentiable functuon defined for all real numbers.

If the following are true:

Which function could be ?

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 .

Compare your answer with the correct one above

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