practice calculus problems

below is a smattering of different types of problems from across the ap calculus ab curriculum. you need to be familiar with these concepts for the multiple choice and free response sections of the exam. a calculator is not needed for any of these problems.

full solutions are given below.

calculus practice problems

1.

2. find

3. let us take the function

when x = 4, is the function increasing or decreasing? is the function concave up, or concave down?

4. a ball is thrown. the height of the ball in meters at any given time can be found by the function

where t is in seconds.

a) at what moment has the ball reached its highest point?
b) what is the acceleration of the ball at this point?

5. a particle moves in a straight line with the velocity given by the equation

what is the total displacement between 0 seconds and 4 seconds?

6.

find r.

7. find the area enclosed by the graphs of

solutions:

1.

2. if we plug in x=0 into the equation, we get a 0 in the denominator. if the expression had evaluated to a real number, the problem would be done. this is not the case here.
our best approach would be to use l’hopital’s rule. take the derivative of the top and bottom, and then evaluate the integral.

3. to find where a function increases or decreases, we need to take the derivative

at x = 4, this evaluates to 26, so this function is increasing at that point.

to determine concavity, let’s take the second derivative

therefore, the function is concave up. since we could look at the initial equation and see that it is an upward facing parabola, we could answer this part without actually taking the derivative but rather by knowing that a parabola’s concavity is always either upward and downward (depending on which way the parabola faces.)

4. to find the moment the ball reaches it highest point, we are looking for the moment the velocity is zero. to find this, take the derivative at set it to 0.

the ball reaches its highest point at t = 4/10 seconds.

the acceleration of the ball at this point is the double derivative of this equation.

the acceleration is a constant -10 m/s.

5. the displacement of a particle can be found by evaluating the integral between the two times given.

this evaluates to

6.

we need to evaluate the integral

therefore, if we solve this, we find that r = 1.

7.

anytime we are looking for the area under a curve, we are looking for an indefinite integral. however, in this case, we also need to know the end points. this is where the two graphs cross.

we find that x = 0 or x = 1, our endpoints.

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for the following practice problems, calculators are not permitted.

practice problems

1.

what is f(g(2))?

a) -12
b) -8
c) -0
d) 8
e) 12

2. calculate the following integral:

a) -1/9
b) -1/27
c) 0
d) 1/27
e) 4/ 9

3. find the slope of the line tangent to the function f(x) at x = 2

a) 17/16
b) 9/8
c) -9/8
d) 2
e) 1

4.

a) ∞
b) -∞
c) 0
d) -1/2
e) 1/2

5. compute the definite integral below:

a) -/2
b) 0
c) /4
d) /2
e)

6. find dy/dx for f(x):

a)
b)
c)
d)
e)

7.

a) 0
b) -1/5
c) 0
d) 1/5
e) ∞

8. what is the area bounded between the graphs of

a) -4
b) 0
c) -32/3
d) 32/3
e) 32

9.  

find dy/dx

a)

b)

c)

d)

e)

10. where does the following function have a local maximum?

a) x = -3
b) x = 0
c) x = 3
d) x = 9
e) x = 27

solutions

1.first, we evaluate g(2) to get -2/(4+1) = -2/5
plugging this into f(x), we can evaluate f(-2/5) = 5*(-2/5)-10 = -12

answer: a) –12

2.this is a fairly straightforward integral that can be solved using the power rule. it’s easier to think it as

answer: e) 4/9

3. anytime we need to find the slope, our first thought should be ‘find the derivative of the function.’

plug in x = 2 to find the slope:

answer: a) 17/16

4. the solution to this question does not involve any calculation. anytime we are asked to evaluate a limit as x approaches infinity, look at the degree of the top and bottom equations. if they’re different, our solution is already given. since in this case, the denominator has a larger degree equation, the limit approaches 0.

answer: c) 0

5. this is one of our common derivatives/integrals which we should know.

answer: c)

6. this is the quotient rule:

in our case:

answer: c) (this can also be reduced to a simpler form.)

7. in this case, the denominator and numerator both evaluate to 0. however, if we factor the top and bottom, we notice that a term cancels out

answer: b) -1/5

8. when we see ‘area bounded by,’ we need to take a definite integral between the points where the graphs intersect. in this case, in since the second equation is y = 0, our limits are the roots of our first equation, x = -2, or x = 2



answer: d) 32/3

9. this is an implicit differentiation problem.

solving for dy/dx

we can reduce this to

answer: b)

10. anytime we are looking for a local maximum, our first step is to take a look at the derivate.

solving for the root of this, we get

the equation has a min and a max at x = 3, and x = -3. you know the max is at x = -3, because if we plug x = -4 into the equation for f’(x), we get a positive answer, so we know the function is increasing before that point.

answer: c) x =3

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at minimum, you should know how to use:

• the power rule
• the chain rule
• the product rule
• the quotient rule

the chain rule is one of the most important rules in this list because many function cans be thought of as functions of functions (of functions of functions).

what is the chain rule and when do we use it?

the chain rule is used when we have a function which is of the form f(x) = g(h(x)).

for example:

our function f(x) is made up of two smaller functions.  you can think of f(x) = g(h(x)), where h(x) = x² and g(x) = sin(x).  we can find the derivative of each of these functions individually.  the chain rule allows us to take the derivative of the entire thing.

the chain rule

in our above example, f(x) = sin(x²).   h(x) = x² , h’(x) = 2x, g(x) = sin(x), and g’(x) = cos(x).  the derivative of h(x) can be solved with the power rule, and the derivative of g(x) is a common derivative.  you can find a list of common derivatives as well as explanations of the other derivative rules in our review of derivative rules.

the difficulty in using the chain rule:

implementing the chain rule is usually not difficult.  the problem that many students have trouble with is trying to figure out which parts of the function are within other functions (i.e., in the above example, which part if g(x) and which part is h(x).
this is going to be a rule which is very important to practice with over and over again.

a handful of example problems

here is a list of example problems of varying difficulties.  you do not need a calculator to answer any of these questions.  you’ll find the solutions below.

solutions and explanations to example problems

 1.

solution:

2.

solution:

3.

solution:

4.

this one is thrown in purposely, even though it is not a chain rule problem.  many students get confused between when to use the chain rule (when you have a function of a function), and when to use the product rule (when you have a function multiplied by a function).

solution:

(this can be reduced to cos(2x) if you know your trig identities.)

5.

we can also use the chain rule in combination with the other rules.

solution:

6.

solution:

7.

again, we can use chain rule in combination with other rules.

solution:

8.

we can use the chain rule multiple times in a row as well.

solution:

9. 

solution:

10. 

solution:

f'(x) = 5(x + 2)⁴

11. f(x) = (x + 2)²

solution:

f'(x) = 2(x + 2)

12.

you will find that chain rule problems often involve sin, cos, and tangent, as we often use trigonometric along with other functions.

solution:

mastering the chain rule is incredibly important for success on the ap calculus exam.  the key to studying the chain rule, as well as any of the differentiation rules, is to practice with it as much as possible.

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ap calculus review: related rates

there is a series of steps that generally point us in the direction of a solution to related rates problems. no two problems are exactly the same, but these steps are a very good rubric for solving a wide variety of problems:

1. we must first identify the variables which are changing in the problem. this could be size, volume, distance, etc.
2. find the governing equation which relates the variables. this is often given in the problem, or is a relatively well-known relation (i.e., volume = length × width)
3. rates are usually (for ap calculus) in relation to time. therefore, we differentiate both sides with respect to time.
4. substitute in the known rates of change and/or known quantities
5. solve for wanted rate of change or quantity

common errors:

we always plug in known values of variables after finding the derivative, never before finding the derivative.

a difficult related rates example.

(note: this is a more difficult than normal problems in its set up. we have simpler sample problems following.)

a ladder with a length of 1 meter is leaned up against a building. it’s a bit off balance, and so beings to slide away from the building at a rate of r m/s. what is the rate that the top of the ladder moves while sliding down the building when the base of the ladder is b meters from the building?

1. variables: what changes with time?
a. speed of the top of the ladder
b. distance of base from the building

2. governing equation:
a. drawing this problem makes it easy to visualize.

b. the moment we see a right triangle, we use the pythagorean theorem
c.
d. we are looking for dy/dt

3. differentiate with respect to t:

we should automatically see at this point that since l is constant, dl/dt is 0. i will leave it for a few steps for purpose of demonstration

4. we can now substitute in what we have.
a. we know dx/dt = r, dl/dt = 0, and x = b

b. we know also know y from the original pythagorean theorem.

which are all known quantities.

related rate example problems

volume questions are quite common examples of related rate problems.

a spherical balloon is being inflated so that its volume increases at a rate of 20 cm3/s. how fast is the radius of the balloon changing when the radius is 10 cm?

solution:

1. identify variables:
volume
radius

2. governing equations

3. we differentiate both sides with respect to time.

4. substitute in the known rates quantities

5. solve for wanted rate of change or quantity

another similar practice problem:

a spherical balloon’s area is increasing at the constant rate of 5 cm/sec. how fast is the volume increasing when the radius is 10 cm?

solution:
this is a two-part question.
first we compute change in r, and then computer the equivalent change in v.

let’s use the same series of steps as before:

part 1:

part 2:

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here is a handful of ap calculus free response style practice problems that you can work through, along with the full solutions. these only represent a small fraction of the types of questions that might come up on an ap calculus ab exam; the more problems you can do before the exam, the better prepared you will be.

ap calculus ab exam free response practice problems

1. an object moves back and forth across the x-axis. its velocity can be modeled by the equation . at time t=10, the car is at a position x = 5. [calculator required]
a) what is the total displacement of the car between the times t = 0 and t=5
b) at time t = 10, is the car slowing down, or speeding up?
c) what is the position of the particle at the initial time?
d) find a time when the velocity equal to 0.

solution:
a) whenever we have a velocity question, we should think back to the following relationships:

the total displacement therefore in the integral of the velocity function between the bounds given

this is an integral that can easily be done by hand. however you should learn how to use you graphing calculator to do this question because many integrals will be far more difficult.

b) slowing down and speeding up are questions about whether or not the acceleration is positive or negative

cos(t) is negative when t = 10. therefore, the object is slowing down.

c) now we need to take the indefinite integral to get a function of position. do not forget to solve for our constant.

to solve for c, we need to know one point: at time t=10, the car is at a position x = 5.

the initial time, t=0 gives us a position: x(0) = 0-cos(0)-.839 = -1.839

d) simply find a root for our initial equation for velocity. t = 5.76 is one of several possibilities.

2. below is the graph of the g(x). [no calculator]

if g(x) = f'(x)
a) where does f(x) have a relative minimum?
b) where does f(x) have a relative maximum?
c) if f(x) has an inflection point, approximately where does it lie?

solution:

a) this function has a relative minimum at x = 0. we should see that it is a minimum because the derivative goes from negative to positive at this point.
b) this function has a relative maximum at x = -2.5 we should see that it is a maximum because the derivative goes from positive to negative at this point.
c) the inflection points exist where the second derivative is 0. we should see this in the graph as existing around x= – 1.25, x = 2, and x = 4

3. we need to enclose an area within a fence. we have 50 feet of fencing available, and a permanent straight wall of infinite length exists on one side (therefore not needing fencing). what dimensions of the fence give the largest possible area?

solution:

this is a quintessential optimization problem. it is often a good idea to sketch the situation given. visualizing the problem often helps in finding the correct governing equations. ap exams sometimes do this for you.

the variables in this problem are the lengths of x and y. these are the only two properties which change.

optimization problem will almost always have two functions involved; in this case the two functions involve area and perimeter.

we know that perimeter p = 50

we need to maximize a = xy with the constraint 50 = x + 2y.

we can rewrite the constraint as one function of another. we put this into the area equation to give:

at this point, we should go back to the constraint and figure out the domain of y. in this case, with 50m of fence, y=25m. we will come back to this later.

we are asked to find the maximum value of a. as in all maximum/minimum value problems, we find the derivative.

we set this equal to 0 and solve for y. in this case, y = 12.5. if we put these numbers back into the constraint:

50 = x + 2y

x = 25

answer:
we are asked for dimensions — the fence is 25m x 12.5m

4. let’s say we want to make a three-dimensional box with a square base (l=w), but we want the surface area to be no greater than 10 m^3. what are the dimensions for a box with the maximum volume?

solution:

this problem is an optimization problem, similar to the last one. if we’re ever not sure, first step is always to draw out the situation:

two governing equations:

equation 1:

equation 2:

we know w = l, so we can use this to simply the first two equations:

we now have a two variable problem:

plug into volume equation

any maximization or minimization, we are going to take the derivative and set to zero. in this case, take derivative of v and set to 0.

we can throw out the negative case because it has no physical meaning in this problem.  always remember, if a question is asking something about the physical world, ask yourself the question ‘does the answer make sense?’  if you get answers that seem ridiculous, go back over your work.

the box is a cube.

final thoughts

these four questions represent just a small fraction of the possible types of questions that can be asked on the ap exam. check back later to see additional sample multiple choice questions, related rates questions, as well as comprehensive reviews of a ton of other topics that you might see on the ap calculus exam.

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understanding the definition of derivatives

before we memorize the derivative rules, when studying for the ap calculus exam, first make sure that you understand the definition of a derivative from the very beginning. questions will certainly arise on the exam which ask you a bit about what to expect will happen to the slope between two points, as those points get closer together on a curve. the definition of a derivative:

you should be able to clearly explain what is happening in this equation, why it is important, and how to use it.

there is a jump here between our definition, and how we often determine derivatives. by the time you get to the ap exam, you need to be very comfortable finding derivatives with the power rule, product rule, quotient rule, and chain rule. as well, you should know a few of the most common derivatives (the 6 trig functions, exponential functions, and logarithm functions). you do not need know how these formulas are created from the definition of the derivative given above, but it is certainly a good exercise to do in your calculus class.

on the ap exam, the derivative is used for many different types of problems.

some of the most common derivative questions will involve:

1. determining the equation of a line tangent to a curve at a point.
2. determining if a function is increasing or decreasing at a given point.
3. determine the critical points of a graph.
4. determining rate of change of physical systems. this involves understanding how velocity
   and acceleration of a particle is related to the position of that particle changing with time.
5. determining the minimum and maximum of a function.
6. using l’hopital’s rule to determine limits of functions (ex: find the lim as x goes to 0 of sin(x)/x)

derivative rules

constant rule:

rule of sums:

rule of differences:

product rule:

quotient rule:

this can also be thought of as the product rule with 1/g(x)

power rule:

chain rule:

examples of derivative rules use:

this is two functions multiplied by each other, so we use the product rule:

this is a function of a function, so we use the chain rule:

this is a division of two functions, so we we look toward the quotient rule:

notice, this is the same thing as saying , which we can solve with the product rule.

common derivatives

common trig derivatives

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important exponent and logarithm rules for ap calculus

below is a list of exponent and logarithm rules with which you should be familiar.

exponent rules

exponent and logarithm practice problems

some common questions on the ap calculus exam involve exponential growth and decay. the growth problems often come up in the form of interest rate questions, where the growth problems often appear in the form of half-life questions. you should be comfortable coming up with an equation to model the following three scenarios (answers below:

a) a bank gives interest at a rate of 4% per year. if the interest in compounded quarterly and one begins with $200, write a formula that will give the total balance in an account as a function of time.

b) in the above question, interest was compounded continuously. write a formula that will give the total balance as a function of time.

c) if the half-life of carbon-14 is 5730 years, write a formula that will give the total percentage of an initial amount remaining as a function of time.

below is a list of problems that you can use to practice your algebra before the ap exam. they involve the above exponent rules, logarithm rules, and quickly changing between bases.

exponent and logarithm practice problems solutions

scenarios

a) t = 200*(1+.04/4)^4t

b) t= 200(e)^.04t

c) percent = (1/2)^t/5730

algebra practice

x = 10/3

x = -2

x = 6/17

x = 7/2

x = 6

x = 3/2

x = 0

x = 2

x =0

x = 9/3

x = 0

using your calculator:

you should be able to solve logarithm and exponential questions quickly on a calculator. remember, when you see a log function on your calculator, it is assumed to be base 10, unless otherwise specified. some calculators (including most graphing calculators) are able to solve log functions which do not have a base 10, but this is not all models. if your calculator cannot do this, or if it is an option buried behind menus and difficult to get to (which is also the case in many graphing calculators), there is a short cut to answering these problems quickly: the change of base formula or…

this means that if we are looking for log₅10, we can plug into our calculator (log 10)/(log 5) and get the correct answer.

being fluent with the algebra we learned before calculus, including little tricks like the change of base formula, can save us a lot of time on the ap exam and allow us time to focus on the more difficult parts of the test.

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the graph shows us that the equation indeed has 2 roots, but we are still not sure what these roots are (although our graphing calculator can solve this for us; see our post on calculator strategies for the ap calculus exam for more).

newton’s method is an iterative method to find approximate roots of equations.

newton’s method usually does not give the exact answer, but will allow us to find very exact approximations. the fist requirement for newton’s method is that we know the derivative of the function.

let’s walk through an example to show where newton’s method comes from.

first step: take a random guess as to what the root might be. let us choose x = 2 for this first guess. we’ll call this x₀. depending on our initial guess, our method might find the first or the second root.

second step: find the equation of a line tangent to the curve at the point x₀.

notice our tangent line has its own root close to the root of our original equation. it is easy to find the root of a linear function. if we take the root of y = 22x-37, we get 37/22, which is about 1.682. this isn’t a perfect approximation, but it’s close. if we were to repeat this entire method, using x = 1.682 instead of x =2, we would get an even closer approximation.

now there is a quick formula that you can derive that gives us our sequence of increasingly accurate approximations.

plugging in x₀ = 2, we get x₁ = 1.682, exactly what we found above. if we do this a few times, we see that we get increasingly close to our root:

x₀ = 1.682
x₁ = 1.51
x₂ = 1.454
x₃ = 1.448

each step brings us closer and closer to the root. we’ll never get perfectly there in this example, but within 4 steps, we’ve within .001. a few more steps, and we’d be within millionths of the correct answer.

newton’s method is an extremely efficient way of finding approximate roots to equations. it works even if the equation is incredibly complicated or would be impossible or difficult to algebraically find exact roots. it rarely gets an exact correct answer, but allows us to get very close. numerical methods, such as newton’s method, for finding roots are the way many computer programs (including many graphing calculators) find answers to equations. the requirement for newton’s method is that you know the derivative of the function.

now let’s practice:

take f(x) = x² – 9. you know the answer to this equation is +/- 3. try newton’s method with this equation to see how many iterations it takes to get within a few thousands of the correct answer.

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reading a derivative graph is an important part of the ap calculus curriculum. typical calculus problems involve being given function or a graph of a function, and finding information about inflection points, slope, concavity, or existence of a derivative.

does the derivative exist?

firstly, looking at a graph we should be able to know whether or not a derivative of the function exists at all. our derivative blog post has a bit more information on this.

the three situations where a derivative does not exist

there is no derivative if there is a discontinuity on the curve.

this is any time that there is a break in a curve, where two parts of the curve do not connect.

types of discontinuity:

there is a removable discontinuity. imagine a linear function such as y = x + 3. if we were to add in a restriction where x is not defined at x = 0, we would have such a discontinuity.

there is an infinite discontinuity. this occurs when we have any equation where there is a break between two continuous sections of a curve due to asymptotes reaching infinity. for example, let y = 3/(x-2). notice, we have two vertical asymptotes that do not connect.

lastly, we have a jump discontinuity. this happens with piecewise functions where two sections simply do not connect.

a derivative does not exist where there is a sharp corner.

this often occurs with absolute value problems. let us look at the graph of y = √x²

at x = 0, there is no derivative because we have a sharp bend in the curve.

lastly, there is no derivative anywhere there is a vertical section of graph.

if there is a vertical section of a graph, the slope is undefined; therefore, the derivative does not exist.

reading the derivative graph.

looking at a graph, we should be able to quickly eye the slope at any and get a rough idea of what the slope should be. this makes it easy to match up a graph with its derivative.

looking at the first graph, can you figure out which of the three below is the graph of the derivative?

f'(x):

a

b

c

few keys to getting the correct answer. we should immediately see that this is some sort of trigonometric function. we know the slope of the function is 0 at a handful of points; therefore the graph of the derivative should go through the x-axis at some point. as well, looking at the graph, we should see that this happens somewhere between -2.5 and 0, as well as between 0 and 2.5. this alone is enough to see that the last graph is the correct answer.

graphing a function based on the derivative and the double derivative.

the derivative and the double derivative tells us a few key things about a graph:

(good ap practice: how can we tell whether it’s min or max?)

the following is a graph of the derivative of f(x).

here is the graph of the function. can we see how they correspond?

being able to read graphs of a derivative and knowing what the general shape of the original function should be is an important part of the ap calculus curriculum. make sure you know how to determine inflection points, local minimums and maximums, and where a function is increasing or decreasing.

the post how to compare a graph of a function and its derivative appeared first on magoosh blog | high school.

which calculator should i get?

there are many calculators currently available. you can find older (see ti-83 and ti-84s) versions that have been around for almost 20 years as well as newer versions that have touch screens and can perform much more complex tasks. there are pros and cons to each. the simplicity of the older ones makes them easier to learn with, leading to less wasted time jogging through menus and options. the newer ones can solve fancy integrals and differential equations.

one of the key strategies is to choose one version early in the year, and use the same calculator for the entire year. by the time you get to the ap exam, the last thing you want to be worried about is figuring out the details of your calculator. you want to know the machine inside and out, and make sure that every key stroke is second nature. these calculators can do amazing things, but you do not want to be searching through menus to figure out how to take an inverse of a trigonometric function, or to figure out how to take logarithms, or find the minimum on a graph. every part of the calculator that you might use on the ap exam should be very familiar by the time you get there.

calculator strategies for the ap calculus exam

besides the basic functions that these calculators share with scientific calculators, you should practice working with the graphing functions.

choosing an appropriate window size

when graphing a function, you need to have an idea of what the function should look like. for instance, let us take a basic example. imagine the equation ? = (x – 10)² − 20. you should have an idea that this is a parabola with a vertex of (10, -20), and two roots. however, if you quickly graph this in a standard window on your calculator, you would see something that might be quickly confused with a linear function. depending on what you’re trying to do, this could lead to mistakes or missing solutions.

know how to resize your window both manually and using the automatic window sizing functions on your calculator.

roots and intersections

you should be able to determine the number of roots and the number of real roots a function should have, and where these roots exist.

firstly, the degree of a polynomial tells us the number of roots. let us take y =x³-3x-3. this is a 3rd-degree polynomial, or a function with 3 roots. however, knowing nothing else about this function, we do not know how many real roots it has. imaginary roots always come in pairs, so this function might have 3 real roots, and 0 imaginary roots, or 1 real root, and 2 imaginary roots, but it cannot have 3 imaginary roots. the graph should be able to tell us more:

notice the graph has a local maximum below the y-axis. since the slope of the curve changes direction twice without crossing the x-axis, we know complex roots exist.

here we have the function y = x³-3x-1. we can see the function cross the x-axis 3 times, and therefore we know we have 3 real roots.

different calculators have different ways of finding these roots. make sure you are able to use your calculator to find the roots of a function.

minimums and maximums

you should be able to calculate the minimums and maximums of graphed functions. different calculators do this in different ways. as well, the shape of the graph should tell you if the point you found is a local or absolute minimum or maximum.

if we look at the example above, y = x³-3x-1, we can see that both x=-1 and x=1 produce a local maximum and a local minimum. although since the graph goes to negative infinity on the left, and positive infinity on the right, we know that these are not absolute minimums and maximums (an odd degree polynomial always has a range of negative infinity to positive infinity).

solving multiple equations

sometimes it is necessary to solve for the solution set of multiple equations. there are many ways of doing this on graphing calculators (some modern calculators have programs built in for this), but often the easiest way is to graph the two (or more) equations and find the intersection. note, this only works for 2 variable equations.

let us find the solution to:

graphing the two equations gives us

we see that there are two intersection points. your calculator will be able to easily find what these two points are, giving us the solution set to our system of equations.

graphing calculators are invaluable tools. different models have a wide variety of abilities. you can easily do well on the ap exam with the simplest of graphing calculators, but the important thing to remember is that you need to know what your calculator is able to do, and how to do it before you get to the ap exam. the exam itself is not the time to be searching through menus on your calculator or wondering if it will be able to solve an equation. by the time the exam comes around, everything you plan on using your calculator for should be second nature.

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