Graphing Functions - AP Calculus AB
Card 0 of 3836
On what intervals does f(x) = (1/3)x3 + 2.5x2 – 14x + 25 increase?
On what intervals does f(x) = (1/3)x3 + 2.5x2 – 14x + 25 increase?
We will use the tangent line slope to ascertain the increasing / decreasing of f(x). To this end, let us begin by taking the first derivative of f(x):
f'(x) = x2 + 5x – 14
Solve for the potential relative maxima and minima by setting f'(x) to 0 and solving:
x2 + 5x – 14 = 0; (x – 2)(x + 7) = 0
Potential relative maxima / minima: x = 2, x = –7
We must test the following intervals: (–∞, –7), (–7, 2), (2, ∞)
f'(–10) = 100 – 50 – 14 = 36
f'(0) = –14
f'(10) = 100 + 50 – 14 = 136
Therefore, the equation increases on (–∞, –7) and (2, ∞)
We will use the tangent line slope to ascertain the increasing / decreasing of f(x). To this end, let us begin by taking the first derivative of f(x):
f'(x) = x2 + 5x – 14
Solve for the potential relative maxima and minima by setting f'(x) to 0 and solving:
x2 + 5x – 14 = 0; (x – 2)(x + 7) = 0
Potential relative maxima / minima: x = 2, x = –7
We must test the following intervals: (–∞, –7), (–7, 2), (2, ∞)
f'(–10) = 100 – 50 – 14 = 36
f'(0) = –14
f'(10) = 100 + 50 – 14 = 136
Therefore, the equation increases on (–∞, –7) and (2, ∞)
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The picture below shows the graph of some function, 
On which interval of 
is the function decreasing?
The picture below shows the graph of some function,
On which interval of
A function is decreasing when it has a negative slope. Graphically, this is a region of the curve where the curve decreases as 
increases.
On interval D, the curve shows this trait. The curve does not show this trait on any other interval.
Therefore, interval D is the answer.
A function is decreasing when it has a negative slope. Graphically, this is a region of the curve where the curve decreases as
On interval D, the curve shows this trait. The curve does not show this trait on any other interval.
Therefore, interval D is the answer.
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Find the intervals that are concave down in 
between the range of 
.
Find the intervals that are concave down in
To find which interval is concave down, find the second derivative of the function.
Now, find which 
values in the interval specified make 
. In this case, 
and 
.
Now to find which interval is concave down choose any value in each of the regions
, and 
and plug in those values into 
to see which will give a negative answer, meaning concave down, or a positive answer, meaning concave up.
A test value of 
gives us a 
of 
. This value falls in the range
, meaning that interval is concave down.
To find which interval is concave down, find the second derivative of the function.
Now, find which
Now to find which interval is concave down choose any value in each of the regions
and plug in those values into
A test value of
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Tessie kicks a bean bag into the air. Its height at a given time can be given by the following equation:
At what time will the bean bag reach its maximum height?
Tessie kicks a bean bag into the air. Its height at a given time can be given by the following equation:
At what time will the bean bag reach its maximum height?
To find the time at which the bean bag reaches its maximum height, we need to find the time when the velocity of the bean bag is zero.
We can find this by taking the derivative of the height position function with respect to time:
Setting this equal to 0, we can then solve for t:
To find the time at which the bean bag reaches its maximum height, we need to find the time when the velocity of the bean bag is zero.
We can find this by taking the derivative of the height position function with respect to time:
Setting this equal to 0, we can then solve for t:
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The postion of a feather in a windstorm is given by the following equation:
Determine when an extrema in the feather's position occurs 
and state whether it is a local minima or maxima.
The postion of a feather in a windstorm is given by the following equation:
Determine when an extrema in the feather's position occurs
The first step into finding when the extrema of a function occurs is to take the derivative and set it equal to zero:
To solve the right side of the equation, we'll need to find its roots. we may either use the quadratic formula:
or see that the equation factors into:
Either way, the only nonnegative root is 2.
To see whether this a local minima or maxima, we'll take the second derivative of our equation and plug in this value of 2:
Since the second derivative is positive, we know that this represents a local minima.
The first step into finding when the extrema of a function occurs is to take the derivative and set it equal to zero:
To solve the right side of the equation, we'll need to find its roots. we may either use the quadratic formula:
or see that the equation factors into:
Either way, the only nonnegative root is 2.
To see whether this a local minima or maxima, we'll take the second derivative of our equation and plug in this value of 2:
Since the second derivative is positive, we know that this represents a local minima.
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Function 
gives the velocity of a particle as a function of time.
Which of the following ordered pairs are the coordinates of a critical point of 
?
Function
Which of the following ordered pairs are the coordinates of a critical point of
Recall that a function has critcal points where its first derivative is equal to zero or undefined.
So, given 
, we need to find v'(t)
Where is this function equal to zero? t=0 for one. We can find the others, but we really just need one. Plug in t=0 into our original equation to find the point (0,0) Which is in this case a saddle point.
Recall that a function has critcal points where its first derivative is equal to zero or undefined.
So, given
Where is this function equal to zero? t=0 for one. We can find the others, but we really just need one. Plug in t=0 into our original equation to find the point (0,0) Which is in this case a saddle point.
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At which point does a local maxima appear in the following function?
At which point does a local maxima appear in the following function?
A local max will occur when the function changes from increasing to decreasing. This means that the derivative of the function will change from positive to negative.
First step is to find the derivative.
Find the critical points (when 
is 
or undefined).
Next, find at which of these two values 
changes from positive to negative. Plug in a value in each of the regions into 
.
The regions to be tested are 
,
, and 
.
A value in the first region, such as 
, gives a positive number, and a value in the second range gives a negative number, meaning that 
must be the point where the max occurs.
To find what the 
coordinate of this point, plug in 
in to 
, not 
, to get 
.
A local max will occur when the function changes from increasing to decreasing. This means that the derivative of the function will change from positive to negative.
First step is to find the derivative.
Find the critical points (when
Next, find at which of these two values
The regions to be tested are
A value in the first region, such as
To find what the
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What is the maximum value of the function 
on the interval 
?
What is the maximum value of the function
First, we need to find the critical points of the function by taking 
.
This is the derivative of a polynomial, so you can operate term by term.
This gives us,
.
Solving for 
by factoring, we get
.
This gives us critical values of 0 and 
. Since we are operating on the interval 
, we make sure our endpoints are included and exclude critical values outside this interval. Now we know the maximum could either occur at 
or 
. As the function is decreasing, we know at the max occurs at 
and that that value is 
.
First, we need to find the critical points of the function by taking
This is the derivative of a polynomial, so you can operate term by term.
This gives us,
Solving for
This gives us critical values of 0 and
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What is the 
-coordinate of the maximum for the function 
?
What is the
To find the maximum, you must first find the derivative of the function, which is 
We found this derivative using the power rule which states,
.
Then, you need to set that equal to zero to get your critical points.
It will be helpful to factor it:
.
Then, set each expression to 0 to get those points:
.
Then, put these points on a number line to then test the sign of the slope in between each point.
To the left of 
, plug in a value (say, 0) into the derivative. You get a positive value. In between 
and 1, the value is negative. To the right of 1, the value is positive. Then, to find the max, look for where the signs change from positive to negative.
That is occurring at 
.
To find the maximum, you must first find the derivative of the function, which is
Then, you need to set that equal to zero to get your critical points.
It will be helpful to factor it:
Then, set each expression to 0 to get those points:
Then, put these points on a number line to then test the sign of the slope in between each point.
To the left of
That is occurring at
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Consider the family of curves given by 
with 
. If 
is a local maximum, determine 
and 
.
Consider the family of curves given by
Since a local maximum occurs at 
, this tells us two pieces of information: the derivative of 
must be zero at 
and the point 
must lie on the graph of this function. Hence we must solve the following two equations:
.
From the first equation we get 
or 
.
To find the derivative we apply the quotient rule
.
Solving 
, we get 
. Plugging this into the expression for 
gives 
.
Since a local maximum occurs at
From the first equation we get
To find the derivative we apply the quotient rule
Solving
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Find the local maximum of the function 
on the interval 
.
Find the local maximum of the function
To find the local maximum, we must find where the derivative of the function is equal to 0.
Given that the derivative of the function 
yields 
using the power rule 
. We see the derivative is never zero.
However, we are given a closed interval, and so we must proceed to check the endpoints. By graphing the function, we can see that the endpoint 
is, in fact, a local maximum.
To find the local maximum, we must find where the derivative of the function is equal to 0.
Given that the derivative of the function
However, we are given a closed interval, and so we must proceed to check the endpoints. By graphing the function, we can see that the endpoint
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Find the local maximum of the curve 
.
Find the local maximum of the curve
First rewrite 
:
Use the multiplication rule to take the derivative:
To find the local extrema, set this to 0...
...and solve for 
...
*
* Since we divided by 
, we have to remember that 
is a valid solution
Therefore, we know that we have two potential local extrema: 
and 
.
By plugging these in, we get two potential local extrema: 
and 
. Therefore, we know that the slope is positive between 
and 
. This means that 
can't be a local maximum, leaving only 
as a potential answer.
Next, we can find the slope at 
. It is:
This is negative, meaning that we go from a positive slope to a negative slope at 
, making it a local maximum.
First rewrite
Use the multiplication rule to take the derivative:
To find the local extrema, set this to 0...
...and solve for
* Since we divided by
Therefore, we know that we have two potential local extrema:
By plugging these in, we get two potential local extrema:
Next, we can find the slope at
This is negative, meaning that we go from a positive slope to a negative slope at
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Find the local maximum of the function 
.
Find the local maximum of the function
To find the local maximum, first find the first derivative of the function.
.
Then find all values of x for which the derivative equals 0 or is undefined. The derivative equals 0 when x=0 and is never undefined because the denominator is always greater than 0. Then, by picking points less than and greater than 0, we see that the function is increasing less than 0 and increasing greater than 0.
Therefore, it is a local maximum.
To find the local maximum, first find the first derivative of the function.
Then find all values of x for which the derivative equals 0 or is undefined. The derivative equals 0 when x=0 and is never undefined because the denominator is always greater than 0. Then, by picking points less than and greater than 0, we see that the function is increasing less than 0 and increasing greater than 0.
Therefore, it is a local maximum.
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What is the maximum of 
over the interval 
?
What is the maximum of
To find the maximum of a function, find the first derivative. In order to find the derivative of this fuction use the power rule which states, 
.
Given the function, 
and applying the power rule we find the following derivative.
Check the 
-value at each endpoint and when the first derivative is zero, namely 
The largest value is 
.
To find the maximum of a function, find the first derivative. In order to find the derivative of this fuction use the power rule which states,
Given the function,
Check the
The largest value is
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Find the 
-value where the local maximum occurs on
.
Find the
To find the maximum of a function, find the first derivative. In order to find the derivative of this fuction use the quotient rule which states,
.
Given the function, 
and applying the quotient rule we find the following derivative.
when 
and 
when 
, which indicates that 
has a local maximum at 
.
To find the maximum of a function, find the first derivative. In order to find the derivative of this fuction use the quotient rule which states,
Given the function,
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Given that the equation of a graph is 
find the 
value of the local maximum on this graph.
Given that the equation of a graph is
To find the critical points of the graph, you first must take the derivative of the equation of the graph and set it equal to zero. To take the derivative of this equation, we must use the power rule,
.
We also must remember that the derivative of an constant is 0. The derivative of the equation for this graph comes out to 
. Solving for 
when 
, you find that 
. The tricky part now is to find out whether or not this point is a local maximum or a local minimum. In order to figure this out we will find whether or not the slope is increasing towards this point or decreasing. Remember that the derivative of a graph equation gives the slope of the graph at any given point.
Thus when we plug in 
into the slope equation, we find that the slope has a positive value. This means that the slope is increasing as the graph leaves 
, meaning that this point is a local minimum, We plug in 
into the slope equation and find that the slope is negative, confirming that 
is the local minimum. That means that there is no local maximum on this graph.
To find the critical points of the graph, you first must take the derivative of the equation of the graph and set it equal to zero. To take the derivative of this equation, we must use the power rule,
We also must remember that the derivative of an constant is 0. The derivative of the equation for this graph comes out to
Thus when we plug in
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What type of point is 
in 
?
What type of point is
Even though the first derivative (
) is 
at 
, there is no max or min because the function is increasing on both sides (derivative is positive on both sides). However, the function is changing its concavity (from negative to positive) at 
. We can tell because the second derivative (
) is also 
at 
, and it's going from negative to positive. Hole and asymptote are irrelevant. When the second derivative changes signs around a specific point we call this an inflection point. An inflection point describes a point that changes the concavity of a function.
Even though the first derivative (
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A ball is tossed into the air and its height is determined by the function 
. What is the maximum height of the ball (in meters)?
A ball is tossed into the air and its height is determined by the function
To determine the maximum height of the ball, the first derivative of the function must be found and set equal to zero.
Test values to the left and right to confirm if this point is a maximum.
Because the derivative goes from positive to negative, we know the original function goes from increasing to decreasing and thus this point is a maximum.
Finally, to find the height of the ball at this point, plug in 
into the original function.
To determine the maximum height of the ball, the first derivative of the function must be found and set equal to zero.
Test values to the left and right to confirm if this point is a maximum.
Because the derivative goes from positive to negative, we know the original function goes from increasing to decreasing and thus this point is a maximum.
Finally, to find the height of the ball at this point, plug in
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What are the 
-coordinate of the local maximum on the graph of the function
?
What are the
To find maxima and minima, find the coordinates of the points where the derivative is undefined or equal to zero. The derivative of p(x) is
Next set the derivative equal to zero and solve for x:
Finally we need to test the critical points in the original equation to determine which is a maximum.
Since the value of the function is greatest at x = -3, that is the x-coordinate of the maximum.
To find maxima and minima, find the coordinates of the points where the derivative is undefined or equal to zero. The derivative of p(x) is
Next set the derivative equal to zero and solve for x:
Finally we need to test the critical points in the original equation to determine which is a maximum.
Since the value of the function is greatest at x = -3, that is the x-coordinate of the maximum.
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Given a graph with equation 
, find the local maximum(s) in the graph if there are any.
Given a graph with equation
In order to solve this equation, we must first underestand that by taking the derivative of an equation of a graph and setting it equal to zero, we can find the values of 
where there are critical points. Critical points are either local maxima or minima, in order to figure out which you simply plug in numbers before and after that value of 
in order to see whether or not the slope increases/decreases as is approaches or leaves that 
value.
In order to take the derivative of equation, the power rule must be applied, 
.
Taking the derivative, we find that the equation becomes 
.
Setting the equation equal to zero and solving for 
, we find that the critical points of this graph are 
. In order to determine whether or not these points are maxima or minima, we plug in numbers larger/smaller as well as inbetween the two critical points to find the behavior of the slope.
For this problem I will choose the numbers -5,0 and 1. These numbers are choosen since 
is approximately -4 and 
is approximately 0.25. By choosing numbers that are larger and smaller then the critical points as well as a number inbetween the two critical points, we can see the behavior of the slope throughout the entire graph.
Plugging in those numbers into 
, we find that 
makes 
positive, 
makes 
negative, and 
makes 
positive. This means that the slope is positive approaching 
and negative leaving it. The slope is negative approaching 
and positive leaving it. For a critical point to be the local maxima, its slope must be positive approaching it and negative leaving it, therefore 
is the local maxima in this graph.
In order to solve this equation, we must first underestand that by taking the derivative of an equation of a graph and setting it equal to zero, we can find the values of
In order to take the derivative of equation, the power rule must be applied,
Taking the derivative, we find that the equation becomes
Setting the equation equal to zero and solving for
For this problem I will choose the numbers -5,0 and 1. These numbers are choosen since
Plugging in those numbers into
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