PCAT Exam: Quantitative Reasoning - A Simple Guide To Calculus

Limits
The limit of a function can be described as the output that is approached as the input approaches a certain value.

Written in function notation, the limit of as approaches a is .

As draws near to some value , represented by , then approaches some number .

In the graph of the function , the line is continuous except where .

Because yields an undefined output and a hole in the graph, the function does not exist at this value.

The limit, however, does exist. As the value is approached from the left side, the output is getting very close to 1.

From the right side, as the -value approaches -2, the output also gets close to 1.

Since the function value from both sides approaches 1, then .
One special type of function, the step function , can be used to define right and left-hand limits.

The graph is shown below. The left-hand limit as approaches 1 is .

From the graph, as approaches 1 from the left side, the function approaches 0.

For the right-hand limit, the expression is .

The value for this limit is one.

Since the function does not have the same limit for the left and right side, then the limit does not exist at .

From that same reasoning, the limit does not exist for any integer for this function.


Sometimes, a function approaches infinity as it draws near to a certain -value.

For example, the following graph shows the function .

There is an asymptote at .

The limit as approaches 3, , does not exist.

The right and left-hand side limits at 3 do not approach the same output value.

One approaches positive infinity, and the other approaches negative infinity.

Infinite limits do not satisfy the definition of a limit.

The limit of the function as approaches a number must be equal to a finite value.


Horizontal asymptotes can be found using limits. Horizontal asymptotes are limits as x approaches either or

For example, to find , the graph can be used to see the value of the function as grows larger and larger.

For this example, the limit is , so it has a horizontal asymptote of

In considering, the limits can also be seen on a graphing calculator by plotting the equation .

Then, the table can be brought up. By scrolling up and down, the limit can be found as approaches any value.
Limit laws exist that assist in finding limits of functions.

These properties include multiplying by a constant, , and the addition property, .

Two other properties are the multiplication property, , and the division property, .

These properties are helpful in finding limits of polynomial functions algebraically.

In , the constant and multiplication properties can be used together, and the problem can be rewritten as .

Since this is a continuous function, direct substitution can be used.

The value of 2 is substituted in for x and evaluated as , which yields a limit of .

These properties allow functions to be rewritten so that limits can be calculated.

 

Continuity
To find if a function is continuous, the definition consists of three steps.

These three steps include finding , finding , and finding .

If the limit of a function equals the function value at that point, then the function is continuous at .

For example, the function is continuous everywhere except . is undefined; therefore, the function is discontinuous at 0.

Secondly, to determine if the function is continuous at 2, its function value must equal its limit at 2.

First, .

Then the limit can be found by direct substitution: .

Since these two values are equal, then the function is continuous at .
Differentiability and continuity are related in that if the derivative can be found at , then the function is continuous at .

If the slope of the tangent line can be found at a certain point, then there is no hole or jump in the graph at that point.

Some functions, however, can be continuous while not differentiable at a given point.

An example is the graph of the function .

At the origin, the derivative does not exist, but the function is still continuous.

Points where a function is discontinuous are where a vertical tangent exists and where there is a cusp or corner at a given -value.
 

Derivatives

Finding Derivatives of Algebraic Functions by Means of the Sum and Product, Power Rule, and Applying the Mean Value Theorem
The derivative of a function is found using the limit of the difference quotient:

.

This finds the slope of the tangent line of the given function at a given point.

It is the slope, , as .

The derivative can be denoted in many ways, such as , , or .
The following graph plots a function in black.

The gray line represents a secant line, formed between two chosen points on the graph.

The slope of this line can be found using rise over run.

As these two points get closer to zero, meaning approaches 0, the tangent line is found.

The slope of the tangent line is equal to the limit of the slopes of the secant lines as

The derivative of a function can be found algebraically using the limit definition.

Here is the process for finding the derivative of

Once the derivative function is found, it can be evaluated at any point by substituting that value in for .

Therefore, in this example, )=4.

Using the Chain Rule to Find Derivatives of Composite Functions
Consider the following functions:

and

The composite function can be built from the two functions as .

Consider the following derivatives: , , and

Therefore, it is true that

This example is the foundation of the Chain Rule, which allows for the differentiation of composite functions.

Basically, if is differentiable at and is differentiable at then the composite function is differentiable at and
In other words, within a composite function, there is an “outside” function and an “inside” function.

In the example above, the outside function is y and the inside function is u.

To find the derivative of the composite function, take the derivative of the outside function and evaluate it in terms of the inside function. Then, multiply times the derivative of the inside function.

In the above example, the derivative of the outside function was 4 and the derivative of the inside function was 5, so their product, 20, was the derivative.

The Chain Rule is helpful when taking derivatives that would otherwise require complicated algebra steps.

Consider the following function I

ts derivative could be taken by squaring out the polynomial, collecting like terms, and then applying the Power Rule to each term.

However, this is complicated and involves too many steps.

The Chain Rule could be applied easily.

The outside function is and the inside function is

Therefore, the derivative of the composite function is the derivative of the outside function, evaluated in terms of the inside function, multiplied by the derivative of the inside function.

This results in

Solving Problems by Differentiation 
Derivatives can be used to find the behavior of different functions such as the extrema, concavity, and symmetry.

Given a function , the first derivative is .

This equation describes the slope of the line.

Setting the derivative equal to zero means finding where the slope is zero, and these are potential points in which the function has extreme values. If the first derivative is positive over an interval, the function is increasing over that interval.

If the first derivative is negative over an interval, the function is decreasing over that interval.

Therefore, if the derivative is equal to zero at a point and the function changes from increasing to decreasing, then the function has a minimum at that point.

If the function changes from decreasing to increasing at that point, it is a maximum. The second derivative can be used to define concavity.

If it is positive over an interval, the graph resembles a U and is concave up over that interval.

If the second derivative is negative, the graph is concave down.

For this equation, solving gets , .

Also, the second derivative is 6, which is positive. The graph is concave up and, therefore, has a minimum value at (0,0).

Finding the derivative of a function can be done using the definition as described above, but rules proved via the different quotient can also be used.

A few are listed below.

These rules apply for functions that take the form inside the parenthesis.

For example, the function would use the Power Rule and Constant Multiple Rule.

To find the derivative, the exponent is brought down to be multiplied by the coefficient, and the new exponent is one less than the original.

As an equation, the derivative is .

In relation to real-life problems, the position of a ball that is thrown into the area may be given by the equation .

The position, , can be found for any time, , after the ball is thrown.

To find the initial position, can be substituted into the equation to find .

That position would be 7ft above the ground, which is equal to the constant at the end of the equation.

Finding the derivative of the function would use the Power Rule.

The derivative is .

The derivative of a position function represents the velocity function.

To find the initial velocity, the time can be substituted into the equation.

The initial velocity is found to be 25ft/s – the same as the coefficient of in the position equation.

Taking the derivative of the velocity equation yields the acceleration equation .

This value is the acceleration at which a ball is pulled by gravity to the ground in feet per second squared.

Using Derivative Tests to Find Extrema, Points of Inflection, and Intervals
Rectilinear motion problems involve an object moving in a straight line.

Given a position function which outputs the position of an object given a specific time t, the object’s velocity can be found by taking the derivative of the position function.

Therefore,

Its speed is equal to the absolute value of the velocity function, .

Also, the acceleration of the object can be found by taking the second derivative of the position function. Therefore,

Optimization problems also make use of derivatives. These problems involve finding either the smallest or largest value of a given function.

The Closed Interval Method can be used to find absolute extrema of a function on a closed interval. First, the critical values need to be found on the given interval. They are found by setting the derivative equal to zero and solving, and they can also exist where the derivative is undefined.

Then, the function is evaluated at those points and at the endpoints of the interval.

The largest value is the absolute maximum within that interval, and the smallest value is the absolute minimum within that interval.

Interpreting the Derivatives of Circular Functions and their Inverses
Circular functions are part of the set of trigonometric functions. They have domains that are angles and ranges that are real numbers.

The most widely used circular (trigonometric) functions are sine, cosine, tangent, cosecant, secant, and cotangent.

 

Here are the derivatives of the six trigonometric functions:


The functions are differentiable at every -value in which they are defined.

For example, tangent is not differentiable at

If tangent is evaluated at  the result is undefined because

There is actually an asymptote there on the graph of tangent.

Therefore, is not in the domain of tangent, and therefore the function is not differentiable there.

 

The derivatives of tangent, cotangent, secant, and cosecant can all be derived using the Quotient Rule and the following identities:
 

Pythagorean identity:

Reciprocal identities:

For example:


Also, the inverse trigonometric functions can be differentiated.

Note that the domain of these six functions are real numbers and their outputs are angles.

Here are the derivatives of the six inverse trigonometric functions:


The derivatives of both the trigonometric functions and the inverse trigonometric functions can be used alone or inside rules such as the Product Rule and Quotient Rule.

For instance, consider the following function

: .

The Product Rule would have to be used on the first half of the function and the Quotient Rule would have to be used on the second half.

Therefore, the derivative is:

Trigonometric functions can be used to model simple harmonic motion.

For instance, the position of a body hanging from a spring that is stretched 4 feet from its resting position can be described by the function where t represents time in seconds.

Note that at time and 4 is known as its amplitude.

Its velocity, can be found by taking the first derivative of the position function and its acceleration, can be found by taking the second derivative of the position function.

Therefore,

and

Interpreting the Derivatives of Transcendental Functions
Transcendental functions are functions that basically “transcend” algebra. In other words, they cannot be expressed in terms of algebraic functions.

These functions include exponential functions, logarithmic functions, and their inverses.

Consider the natural logarithmic function and its inverse function, the exponential function

 

The derivative of the exponential function is actually the function itself. It is the only function that holds this property.

Therefore,

.
The derivative of its inverse function, is derived using implicit differentiation.

The rule is that

for
Other rules exist for exponential functions and logarithmic functions with bases other than

Consider the exponential function with base

Its derivative can be derived using the Chain Rule and it can be seen in the rule:


This rule is applied when finding that the derivative of

is
Finally, the logarithmic function with base which is , can be derived using the Change of Base formula for logarithms and the derivative of the natural logarithmic function.

The process is shown here:

for
This rule is applied when finding that the derivative of

is

Determining the Derivatives of Composite Functions
Composite functions are often built from trigonometric, logarithmic, and exponential functions.

In this case, in order to take the derivative, the Chain Rule must be used.

Consider the example of a composite function

It is a composite function with an outside function of and an inside function of

The derivative of the outside function is but evaluating it in terms of the inside function results in

The derivative of the inside function is

Therefore, the derivative of the composite function is the product of these two expressions:

The following function also requires the Chain Rule to find its derivative:


Note that the inside function is a transcendental function and the outside function is a transcendental function

Therefore, its derivative is


The Chain Rule can be used as many times as necessary, depending on how many inside functions exist within a composite function.

Consider the following composite function based on the circular function cosine:

It has an outside function of a middle function of , and an inside function of

 

Its derivative involves using the Chain Rule twice, and is

Using Implicit Differentiation
If an equation is not placed into the form , in order to be differentiated, implicit differentiation must be used.

Basically, if the equation is not solved for as a function of , the normal rules for taking derivatives do not apply.

These equations are known as implicitly defined functions.

Some examples of implicitly defined functions are

and

Solving for as a function of in both examples is very difficult; therefore, in order to differentiate each function, a new technique must be used.
In this case, the process is called implicit differentiation, and it involves taking the derivative of both sides of the equation with respect to . The variable y must be treated as a function of , so when taking the derivative of term, the chain rule must be used. For instance,

Once the derivative of both sides are taken, the equation is then solved for in terms of both and .

This end result gives a formula for the slope of the tangent line at any point on the original curve.

 

A common function that highlights the use of implicit differentiation is one that represents a circle.

Consider the equation a circle centered at the origin with radius 3. Its derivative can be found by using implicit differentiation.

First, take the derivative of both sides with respect to .

Remember, that whenever there is a , the chain rule must be used because is a function of .

This results in

Then, solve for using algebra.

This results in , which is the formula for the slope of the tangent line at any point () on the original circle.

This example also could have been done using normal differentiation by first solving for y to obtain.

The chain rule would be used to obtain the same result.
Knowing the derivative of the natural logarithmic function and how to differentiate implicitly are helpful tools when differentiating functions in which other rules cannot be applied.

The process is known as Logarithmic Differentiation.

Consider the function

There is no known rule that allows the derivative to be taken.

Therefore, take the natural logarithm of both sides and use properties of logarithms to rewrite it as

Differentiating implicitly on both sides with respect to results in

Therefore,

Determining Related Rates
Related rate problems also involve derivatives. Each problem involves both an unknown quantity and known quantities that involve derivatives. The key is to relate the unknown quantity or its rate of change to the known quantities or their rates of change through a known formula or equation. The equation must be differentiated with respect to the independent variable.

The functions usually have time as their independent variables, so most derivatives are with respect to time and implicit differentiation needs to be used.

Consider an object moving along the path and at some time t, its -coordinate is 8 and the -coordinate is moving at a rate of 4 units per measurement of time.

To find how fast the -coordinate is moving, a related rates problem must be solved.

Using implicit differentiation,

The problem gives that and so this equation can be used to find that the -coordinate is moving at a rate of 768 units per measurement of time.

Integrals

 

Finding Antiderivatives and Interpreting C
Per the fundamental theorem of calculus, on a closed interval [a, b], the following represents the definite integral: : . represents the antiderivative of the function

Other theorems allow constants to be moved to the front of the integral, negatives to be moved to the outside of the integral, and integrals to be split into two parts that make up a whole.

An example of using these theorems can be seen in the following problem: .

The antiderivative of is
Within the fundamental theorem of calculus, the antiderivative exists.

It is true that

Therefore, it is important to know how the graph of a function and a derivative relate.

Because the derivative function represents the slope of the tangent of a function, where a function is horizontal, the derivative function has zeros.

On intervals where the function is decreasing, the derivative function lies below the -axis, and on intervals where the function is increasing, the derivative function lies above the y-axis.

Slope is defined in algebra to be a rate of change; therefore, the derivative function is a rate of change.

The definite integral in the fundamental theorem of calculus can also be used to represent a rate of change.

If one were to calculate the definite integral of a function over the interval as  , where the result is the net rate of change of over the same interval.

Understanding and Using Sigma Notation for Simplifying Sums
Sigma notation can be used to simplify writing down long sums.

Suppose that there is a list of numbers, which can be labeled by the numbers 1, 2, 3, and so on.

This label is called an index, and it is generally written at the lower right of the symbol representing the numbers in the list: .

Thus, will indicate the first number in the list, will indicate the second number, and represents the i-th number in the list.

Suppose that for a given problem, it is necessary to plug each number in the list into some formula, f, then add up the results for every number.

The resulting sum would be quite long and difficult to write out, since it would look like and so on.

In addition, it is sometimes necessary to write down formulas that work for different lists that may have a different length.

Sigma notation is a way of writing down such sums that solves this problem elegantly.
Therefore, suppose the list of numbers runs from to , where n is an arbitrary natural number.

To indicate the sum of all n quantities , one may use sigma notation as follows:

The large symbol is a capital Greek letter sigma.

Below it, write the index label (in this case i) and the number at which to start. At the top, write the number at which the sum is to stop.

Then, to the right of the sigma symbol itself, write the expression that is to be added up.
It is not necessary to always start the sum at the first value of the index.

By writing some other number instead of 1, sigma notation can be used to indicate that the sum is to start at some other point in the list.

So, for example, , while .

Approximating Areas Bounded by Curves
In calculus, the area problem involves finding the area under a positive function from to above the -axis.

 

Such a region Ω is shown here:


The area is defined as the definite integral of from to and is denoted as

In a similar manner, the area between two curves is and from to where over that same interval is given as

Arc length can also be calculated using an integral.

Consider the same function from to

The length of the curve over that interval, also known as arc length, is defined as

All three of these integral definitions are based on proofs based on limits.

The average value of a function can be found by the following integral:

.

The integral finds the area of the region bounded by the function and the -axis, while the fraction divides the area to find the average value of the integral.

An example of this is shown in the graph below.

The function is the black line.

The light gray shading represents the area under the curve, while the rectangle drawn on top with added darker shading represents the same amount of area as the region under the graph of the given function over the interval of a to b.

This rectangle has the base [a, b] and height f(c).


The Mean Value Theorem states that if is a continuous function on interval [a, b], and is differentiable on (a, b), then there exists at least one number, c, in which the derivative at that point equals the slope of the secant line connecting the endpoints of the interval.

This number can be found by the equation

.

Integration
Since integration is the inverse operation of finding the derivative, the integral is found by going backwards from the derivative.

In relation to the ball problem, an acceleration function can be integrated to find the velocity function.

That function can then be integrated to find the position function.

From velocity, integration finds the position function , where is an unknown constant. More information would need to be given in the original problem to integrate and find the value of .