Calculus I

7/17/2019 Calculus I

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Review



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Functions

Within this section, we will review the notation and properties of functions, and

their uses.

Definition 1.1

A function f (x1, ..., xn) is a relation between a set of inputs (domain) and a set of outputs

(range). A formula or algorithm may define the relation between the inputs and the outputs.

Functions are considered the basis of modern calculus. They can define complex sets

of points as an equation or algorithm. Within this course, we will manipulate functions

for applications in the real world.

A single variable function takes in an argument x , and returns an output f (x). Func-

tions are often written as an equation which relates f (x) to x . f (x) = 3x2 − 5 is con-sidered a function and is in proper form. In order to be considered a function, for each

x-value, there will be one and only one corresponding f (x) (assuming f (x) won’t be

undefined) value. If an equation maps one x-value to multiple f (x) outputs, it is not

considered a function.

As well as single variable functions, multi-variable functions exist as well. These

work similar to single variable functions, although they are written as f (x1,..., xn) in-

stead.

The notation we will use is primarily f (x) (Which you should NOT confuse with

multiplication). If we are using multiple functions, we will then continue to g(x) forthe next function, h(x) for the next, and so on. Occasionally, when only the function is

manipulated, f alone will be written.

Example 1.2

The polynomial 5x4 + 3x2 − x + 5 is a function.

This polynomial is considered a function, because all inputs can only be mapped to

one output. To write it as a function, we would say f (x) = 5x4 + 3x2 − x + 5.

Example 1.3

Is the polynomial 27x5 − 10x3 + 2x2 − 5x a function?

For this problem, we can simply see all terms are real and only have one output,

meaning the sum of all terms (the polynomial itself) has to be real and only have one

output.

The roots of a function are values of x which satisfy the equation f (x) = 0. Solving

for roots describes all points at which the function hits the x-axis. One can modify thisequation in a way that describes the intersections of a line parallel to the x-axis. This



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modification is written as f (x) = n, which solves all points at which the function is

equal to n.

Consider the polynomial x3 − 4x2 − 7x + 10. The roots of the equation can be solved

by factoring and using the factor theorem. The factor theorem states if x − α is a factor

of a polynomial P then α is a root of P , and therefore satisfies the equation P (α) = 0.Using this method, we end up with the factored equation

(x − 5)(x − 1)(x + 2) = 0 (1.1)

which yields roots 5, 1, and −2. We can check these roots graphically by graphing

the function; that is, graphing every point with coordinates (x, f (x)).

Example 1.4

Find the roots of function f (x) = x2 − 16

To find the roots, we first must consider f is a polynomial. Because it is a polyno-

mial, we can solve it using the factoring method we described earlier. First, we factor

the equation to get

(x − 4)(x + 4) = 0 (1.2)

which, using the factor theorem, yields roots 4 and

−4.

Example 1.5

Find the roots of function f (x) = 2x.

It is generally known that all exponential functions have horizontal asymptotes; that

is, they approach 0 on the x-axis but never reach it. This means there will be no roots,

as roots are defined as the points which a function intersects the x-axis. Nevertheless,

we can try and solve it algebraically.

For this function, we can first take the logarithm of each side in the roots equation,

giving log2(2x) = log2(0). For the first side, the answer is obviously x, while for the

second side, we see that it is undefined; logarithms with base b ≤ 0 are considered

undefined.

We can try and avoid this by making the second side greater than 0 for the logarithm

to work. By adding 2x to each side, we get 2 · 2x = 2x. With this equation, we take the

logarithm (of base 2) on each side and simplify it to get

log2 2x = log2(2 · 2x) (1.3)

x = log2 2 + log2 2x

(1.4)x = x + 1 (1.5)



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Here we can clearly see that x equals something other than itself, making it unde-

fined.

There is no way to solve for the roots of any exponential function.

For some functions, finding the roots is as simple as solving an elementary algebraicequation. However, for some functions, like polynomials, solving can require more

complex processes. Other functions, like sin and cos can have infinite solutions, pro-

vided there are no extra terms.

Most of the time, you can approximate the roots of the function by graphing it and

checking points which the function intersects the x-axis. However, this is generally in-

accurate and should only be used as a way to check your solutions already calculated

algebraically.

Definition 1.6The composite function of f and g, written f ◦ g, is the pointwise application of f on g,

equal to f ( g(x))

Function composition allows for easier notation, but won’t generally be useful for

this course. However, it is important to note this notation in case it is used.

Function iteration, i.e. applying function compositions multiple times upon itself can

be written as a superscript above f . For example, f ◦ f ◦ f = f 3(x). It is also useful to

note the function f 0(x) is equal to x.

Function iteration also works with various exponential rules. For example, the prod-uct rule with function iteration, f x f y = f x+ y works for any function f . The power rule,

f x y

= f xy also works.

f −1(x) is the inverse function of f . The inverse function of f is a mapping Y → X

which takes in a value of f (x) and returns all x values which satisfy it.

A simple trick to invert a function is to switch all x values with f (x), then solve for

f (x).

Example 1.7

Evaluate the inverse function f −1 of f (x) = x2.

To solve this problem, we only need to switch one variable. This makes the equation

f (x)2 = x, which when square rooted gives the function f −1(x) =√

x.

Example 1.8

Evaluate the inverse function f −1 of f (x) = 5x−7x+6 .

For this rational function, there are two variables we need to switch; one in the nu-merator, one in the denominator. By substituting we get



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x = 5 f (x)− 7

f (x) + 6 (1.6)

Which we can then simplify (using many algebraic properties) to

f −1(x) = −6x − 7

x − 5 (1.7)

The inverse functions of rational functions are often tedious to calculate, and require

checking, which we will learn how to do in the next section. In that section, we will

focus specifically on inverse functions and their properties.

At the end of each section in this book, there will be various exercises to challenge

your thinking. Although not required, it is recommended to do so for better review.

Exercises

(1) Is y/x = 3 a function?

(2) Is Re(eix ) a function?

(3) Graphically solve for all roots of f (x) = x2 − 1 and f 2(x).

(4) Solve for all roots of f (x) = x3 − 2x2 − 3x.

(5) Solve for all roots of f (x) = x2 + 7x + 10.

(6) Solve for all roots of f (x) = 2x2 − x − 6.

(7) What is f ◦ g for g = f 2?

(8) What is the inverse function of x2?

(9) Name a function that will not change with any function iteration.

(10) Create an equation that describes the inverse function of any exponential func-

tion xn.



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Inverse Functions

Within this section, we will review inverse functions, their uses, applications, and

properties.

As mentioned in the first section, the easiest way to solve for an inverse function isto switch the variables of the function, and solve for the value of the function. This

method will be used as it is simple, easy, and accurate.

We will not check algebraically for this section, but during exams, to check alge-

braically, you must take the inverse of your answer. If you end up with the original

function, your inverse is correct. This method works because inversion applied twice

will not affect the function.

Example 1.9

Evaluate the inverse function f −1 of f (x) = 3/x + 2.

For this function, we can switch the variables to get

x = 3

f −1 + 2 (1.8)

f −1x + 2x = 3 (1.9)

f −1x = −2x + 3 (1.10)

f −1

(x) = −2x + 3

x (1.11)

Example 1.10

Evaluate the inverse function f −1 of f (x) = 2x.

For this function, we can switch the variables and take the logarithm of base 2 on

each side to get

f −1 = log2 x (1.12)

This works well with the definition of the inverse exponential function as a loga-

rithm, and the definition of an exponent as an "anti-logarithm".

Example 1.11

Evaluate the inverse function f −1 of linear function f (x) = ax + b.

For this function, we can again, switch the variables and simplify to get

x = a f −1 + b (1.13)



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x − b = a f −1 (1.14)

f −1 = x − b

a (1.15)

Example 1.12

Evaluate the inverse function f −1 of f (x) = x2.

For this function, we can switch the variables, and square root each side to get f −1(x) =√ x.

Example 1.13

Evaluate and graph the inverse function f −1 of f (x) = 4/x + 9.

For this function, we can switch the variables, and graph both f and f −1

x = 4

f −1 + 9 (1.16)

f −1x + 9x = 4 (1.17)

f −1x = −9x + 4 (1.18)

f −1(x) = −9x + 4

x (1.19)

Now that we have defined both f and f −1, we can graph them.

Along with f and f −1, the identity function idX is also drawn (in purple) on the

graph.

It is drawn to display a very noticeable feature. The inverse function is a reflectionover the identity function idX . This feature must be present in every single inverse



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function. We can use this to our advantage as an easy way to check our answers graph-

ically without using the algebraic method.

Example 1.14

Evaluate, graph, and check the inverse function f −1 of f (x) = ex.

Recall that the inverse of an exponential equation is a logarithm, and base of this ex-

ponent is e. This means the logarithm’s base must be e, making it the natural logarithm

function ln.

Now that we have f and f −1, we can graph it and check if it is a reflection along the

identity function.

It’s obvious that the inverse function is a reflection along the identity function, so

our inverse must be correct.

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