Preliminaries About Functions

Preliminaries About Functions

Noel DeJarnette and Austin Rochford

August 21, 2012

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Symbols

∈, ”in” or ”an element of”, 1ε {0, 1, 2, 3, 4}⊂, ”subset of” or ”contained in”,{0, 1, 2, 3, 4} ⊂ N ⊂ Z ⊂ Q ⊂ R(a, b), interval notation for {x ∈ R : a < x < b}, (−∞,∞) isall of R.

{x ∈ R : x ≥ a}, set notation for [a,∞).

N, natural numbers, {1, 2, 3, 4, . . .}Z, integers, {. . . ,−4,−3,−2,−1, 0, 1, 2, 3, 4, . . .}

Q, rational numbers,{

x ∈ R : x = pq where p, q ∈ Z, q 6= 0

}

Functions, a Doodle

The functions we will deal with in Calc 1 (and probably all thefunctions you have seen and will see until Calc 3) send one realnumber to another real number.

so f takes pts from a set, A, of the real numbers (it could be all ofR

Functions, a Doodle


so f takes pts from a set, A, of the real numbers (it could be all ofR)

Functions, a Doodle


so f takes pts from a set, A, of the real numbers (it could be all ofR)

to another set, B, also in the reals.

Functions, a Doodle


so f takes pts from a set, A, of the real numbers (it could be all ofR) to another set, B, also in the reals.

Functions, a Doodle


More precisely, a number a ∈ A

Functions, a Doodle


More precisely, a number a ∈ A is sent to a point f (a) = b ∈ B

Domain and Range of Functions

Definition

The domain of a function f is the set of points where f isdefined,

(meaning f (x) exists)

.

This would be the set A in the doodle.

Example

The function f (x) = x2 is defined everywhere, so the domain is allof R.

Example

f (x) = ln x is defined only for strictly positive numbers so thedomain is {x ∈ R : x > 0}.


Definition

The domain of a function f is the set of points where f isdefined,(meaning f (x) exists).

This would be the set A in the doodle.

Example

The function f (x) = x2 is defined everywhere, so the domain is allof R.

Example

f (x) = ln x is defined only for strictly positive numbers so thedomain is {x ∈ R : x > 0}.


Definition

The range of a function f is the set of points that are the valuesf (x).

This would be the set B in the doodle.

Example

The function f (x) = x2 is always nonnegative, so the range is{x ∈ R : x ≥ 0}.

Question

What does nonnegative mean? How does the domain of ln xdiffer from the range of x2?

Example

The function f (x) = ln x gives all values from −∞ to ∞ so therange is all of R.


Sometimes we want information about subsets of the range or thedomain.

The Preimage is a perfect example. It will be used to find inversesof functions and is important in finding the domains ofcompositions.


Sometimes we want information about subsets of the range or thedomain.The Preimage is a perfect example. It will be used to find inversesof functions and is important in finding the domains ofcompositions.


The Preimage will be a subset of the domain.

First, look at subset of the range, C.


The Preimage will be a subset of the domain.First, look at subset of the range, C.


The Preimage will be a subset of the domain.First, look at subset of the range, C.The Preimage of C is all the points in A that f sends to C.


The Preimage will be a subset of the domain.First, look at subset of the range, C.The Preimage of C is all the points in A that f sends to C.

If all the points of D are sent to C, and those are the only pointssent to C, then D is the preimage of C


The Preimage will be a subset of the domain.First, look at subset of the range, C.The Preimage of C is all thepoints in A that f sends to C.

Formally, Preimage(C ) = {x ∈ A : f (x) ∈ C}.

Graphs of Functions

This picture might be helpful in understanding what functions do,but it doesn’t really explain how the function behaves.

You aren’t able to deduce much about a function if all you know isthat it sends (0,∞) (0,∞).This is where the graph of the function comes in.

Graphs of Functions

This picture might be helpful in understanding what functions do,but it doesn’t really explain how the function behaves.You aren’t able to deduce much about a function if all you know isthat it sends (0,∞) (0,∞).

This is where the graph of the function comes in.

Graphs of Functions

This picture might be helpful in understanding what functions do,but it doesn’t really explain how the function behaves.You aren’t able to deduce much about a function if all you know isthat it sends (0,∞) (0,∞).This is where the graph of the function comes in.

Graphs of Functions

This picture might be helpful in understanding what functions do,but it doesn’t really explain how the function behaves.We aren’t able to deduce much about a function if all we know isthat it sends (0,∞) (0,∞).This is where the graph of the function comes in.

Graphs of Functions

The graph of a function is a set in the cartesian plane (R2)

The points in the set are the pair (a, f (a), a point in the domainwith its associated point in the range.

Graphs of Functions

The graph of a function is a set in the cartesian plane (R2)The points in the set are the pair (a, f (a), a point in the domainwith its associated point in the range.

Graphs of Functions

The graph of a function is a set in the cartesian plane (R2)The points in the set are the pair (a, f (a), a point in the domainwith its associated point in the range. Plot all the points and wehave an informative way to visually describe a function.

Deducing Domains by Inspection

Let’s begin with examples we have already seen.

If the x-axis represents the inputs and the y-axis represents thevalues of f (x), or the outputsWe can easily see that the range is all values greater than or equalto 0. It is harder to see the domain, but since we don’t see anydiscontinuities, we assume it keeps extending, and the domain is allreal numbers.



If the x-axis represents the inputs and the y-axis represents thevalues of f (x), or the outputs

We can easily see that the range is all values greater than or equalto 0. It is harder to see the domain, but since we don’t see anydiscontinuities, we assume it keeps extending, and the domain is allreal numbers.



If the x-axis represents the inputs and the y-axis represents thevalues of f (x), or the outputsWe can easily see that the range is all values greater than or equalto 0.

It is harder to see the domain, but since we don’t see anydiscontinuities, we assume it keeps extending, and the domain is allreal numbers.



If the x-axis represents the inputs and the y-axis represents thevalues of f (x), or the outputsWe can easily see that the range is all values greater than or equalto 0. It is harder to see the domain, but since we don’t see anydiscontinuities, we assume it keeps extending, and the domain is allreal numbers.

Aside about Polynomials

We just saw that the domain of f (x) = x2 is R.

We can easilysame for f (x) = −x − 1

In fact all polynomials will have a domain of all real numbers.

Question

What polynomials will have a range of all real numbers?

Aside about Polynomials

We just saw that the domain of f (x) = x2 is R. We can easilysame for f (x) = −x − 1

In fact all polynomials will have a domain of all real numbers.

Question

What polynomials will have a range of all real numbers?


Let’s look at a function whose domain is not all of R.

We can see there are no values for the function to the left of zero.The domain is {x ∈ R : x ≥ 0}. Same for the range.

Question

Can you find a function whose domain is not all real numbers, butwhose range is?



We can see there are no values for the function to the left of zero.The domain is {x ∈ R : x ≥ 0}.

Same for the range.

Question




We can see there are no values for the function to the left of zero.The domain is {x ∈ R : x ≥ 0}. Same for the range.

Question


Domains of Combined Functions

We can add, subtract, multiply, and divide functions just as we cannumbers.

After all, the outputs of functions we have seen so far are justelements in the real numbers.

Question

What is the domain of h(x) = f (x) + g(x) in terms of the domainsof f and g?

Both f (x) AND g(x) need to be defined for h(x) to exist.Since both need to be defined we have the domain of h(x) =domain f

⋂domain g .

Question

How would you have to adjust the domain for h(x) = f (x)g(x)?


We can add, subtract, multiply, and divide functions just as we cannumbers.After all, the outputs of functions we have seen so far are justelements in the real numbers.

Question



⋂domain g .

Question




Question


Both f (x) AND g(x) need to be defined for h(x) to exist.

Since both need to be defined we have the domain of h(x) =domain f

⋂domain g .

Question




Question



⋂domain g .

Question


Domains of Compositions of Functions

Compositions of functions is simply where we use one function asthe input for another function.

This is most helpful when exploring translations and reflections ofgraphs of functions, but it will come up in many other contexts.

Example

Let f (x) = ln x and g(x) = x2, thenh(x) = g ◦ f (x) = g(f (x)) = g(ln x) = (ln x)2 = ln2 x .

Question

What conditions must be met for h(x) to exist? (i.e. what is thedomain of h(x))


Compositions of functions is simply where we use one function asthe input for another function.This is most helpful when exploring translations and reflections ofgraphs of functions, but it will come up in many other contexts.

Example

Let f (x) = ln x and g(x) = x2, thenh(x) = g ◦ f (x) = g(f (x)) = g(ln x) = (ln x)2 = ln2 x .

Question

What conditions must be met for h(x) to exist? (i.e. what is thedomain of h(x))


g(x) = x2 is a polynomial, so its domain is all real numbers.

Since the domain is all real numbers, that will clearly encompassthe range of f .We conclude that the domain of the composition, h, is the same asthe domain of f ,{x ∈ R : x > 0}

Question

Which statement is always true for any f and g: domain of f ⊂domain of g ◦ f , domain of g ◦ f ⊂ domain of f ?

Question

What is the domain of f ◦ g(x) = ln x2?


g(x) = x2 is a polynomial, so its domain is all real numbers.Since the domain is all real numbers, that will clearly encompassthe range of f .

We conclude that the domain of the composition, h, is the same asthe domain of f ,{x ∈ R : x > 0}

Question


Question



g(x) = x2 is a polynomial, so its domain is all real numbers.Since the domain is all real numbers, that will clearly encompassthe range of f .We conclude that the domain of the composition, h, is the same asthe domain of f ,{x ∈ R : x > 0}

Question


Question



f (x) = ln x has a domain of {x ∈ R : x > 0}.

Thus f (g(x)) will defined for all x such that g(x) > 0.The range of g is always NONNEGATIVE, which means we haveto only worry about the x ’s where g(x) = 0.So the domain of f ◦ g is {x ∈ R : x 6= 0}We see that domain f ◦ g ⊂ domain g


f (x) = ln x has a domain of {x ∈ R : x > 0}.Thus f (g(x)) will defined for all x such that g(x) > 0.

The range of g is always NONNEGATIVE, which means we haveto only worry about the x ’s where g(x) = 0.So the domain of f ◦ g is {x ∈ R : x 6= 0}We see that domain f ◦ g ⊂ domain g


f (x) = ln x has a domain of {x ∈ R : x > 0}.Thus f (g(x)) will defined for all x such that g(x) > 0.The range of g is always NONNEGATIVE, which means we haveto only worry about the x ’s where g(x) = 0.

So the domain of f ◦ g is {x ∈ R : x 6= 0}We see that domain f ◦ g ⊂ domain g


f (x) = ln x has a domain of {x ∈ R : x > 0}.Thus f (g(x)) will defined for all x such that g(x) > 0.The range of g is always NONNEGATIVE, which means we haveto only worry about the x ’s where g(x) = 0.So the domain of f ◦ g is {x ∈ R : x 6= 0}

We see that domain f ◦ g ⊂ domain g


f (x) = ln x has a domain of {x ∈ R : x > 0}.Thus f (g(x)) will defined for all x such that g(x) > 0.The range of g is always NONNEGATIVE, which means we haveto only worry about the x ’s where g(x) = 0.So the domain of f ◦ g is {x ∈ R : x 6= 0}We see that domain f ◦ g ⊂ domain g

Domains of Compositions of Functions Graphically

We haven’t yet talked about inverses, but we should know that x2

and√

x are inverses, at least philosophically, so√

x2 =√

x2 wherethey are both defined.

We will see that√

x2 and√

x2

will have very different domains.

The Domain of√x2

Let’s first look at√

x2 which you might recognize as the absolutevalue, |x |.

The Domain of√x2

The range of x2 are the nonnegative real numbers.

The Domain of√x2

Which coincides with the domain of√

x .

The Domain of√x2

Thus√

x2 is defined everywhere, (The domain is all real numbers).

The Domain of√x

2

What happens if we compose in the other order?

The Domain of√x

2

Right away we see that the domain of√

x2

will be restricted as thedomain of

√x is x ≥ 0.

The Domain of√x

2

Again, the domain of x2 is all real numbers which contains therange of

√x .

The Domain of√x

2

So we conclude that the domain of√

x2

= {x ∈ R : x > 0} 6=domain of

√x2

Translations via Compositions

It is important that we have the graphs, or at least rough sketches,of basic functions memorized.

We will then use translations to generate a much larger class ofgraphs.If we know what the graph of f (x) looks like, then we will be ableto find the graphs of −f (x), f (−x), f (x + c), f (x) + c where c isany constant.We would like to be able to do this by just transforming the graphand not computing values for the new function.


It is important that we have the graphs, or at least rough sketches,of basic functions memorized.We will then use translations to generate a much larger class ofgraphs.

If we know what the graph of f (x) looks like, then we will be ableto find the graphs of −f (x), f (−x), f (x + c), f (x) + c where c isany constant.We would like to be able to do this by just transforming the graphand not computing values for the new function.


It is important that we have the graphs, or at least rough sketches,of basic functions memorized.We will then use translations to generate a much larger class ofgraphs.If we know what the graph of f (x) looks like, then we will be ableto find the graphs of −f (x), f (−x), f (x + c), f (x) + c where c isany constant.

We would like to be able to do this by just transforming the graphand not computing values for the new function.


It is important that we have the graphs, or at least rough sketches,of basic functions memorized.We will then use translations to generate a much larger class ofgraphs.If we know what the graph of f (x) looks like, then we will be ableto find the graphs of −f (x), f (−x), f (x + c), f (x) + c where c isany constant.We would like to be able to do this by just transforming the graphand not computing values for the new function.

Left and Right Translations

Let f (x) be the function we want to transform and g(x) = x + c .

We will see that f ◦ g = f (x + c) and g ◦ f = f (x) + c will bedifferent functions.Horizontal, or left/right translations are formed by precomposing fwith g , or f (x + c).If c is positive, then the graph of f will be shifted to the left.If c is negative, then the graph of f will be shifted to the right.Let’s look at an example.


Let f (x) be the function we want to transform and g(x) = x + c .We will see that f ◦ g = f (x + c) and g ◦ f = f (x) + c will bedifferent functions.

Horizontal, or left/right translations are formed by precomposing fwith g , or f (x + c).If c is positive, then the graph of f will be shifted to the left.If c is negative, then the graph of f will be shifted to the right.Let’s look at an example.


Let f (x) be the function we want to transform and g(x) = x + c .We will see that f ◦ g = f (x + c) and g ◦ f = f (x) + c will bedifferent functions.Horizontal, or left/right translations are formed by precomposing fwith g , or f (x + c).

If c is positive, then the graph of f will be shifted to the left.If c is negative, then the graph of f will be shifted to the right.Let’s look at an example.


Let f (x) be the function we want to transform and g(x) = x + c .We will see that f ◦ g = f (x + c) and g ◦ f = f (x) + c will bedifferent functions.Horizontal, or left/right translations are formed by precomposing fwith g , or f (x + c).If c is positive, then the graph of f will be shifted to the left.

If c is negative, then the graph of f will be shifted to the right.Let’s look at an example.


Let f (x) be the function we want to transform and g(x) = x + c .We will see that f ◦ g = f (x + c) and g ◦ f = f (x) + c will bedifferent functions.Horizontal, or left/right translations are formed by precomposing fwith g , or f (x + c).If c is positive, then the graph of f will be shifted to the left.If c is negative, then the graph of f will be shifted to the right.

Let’s look at an example.


Let f (x) be the function we want to transform and g(x) = x + c .We will see that f ◦ g = f (x + c) and g ◦ f = f (x) + c will bedifferent functions.Horizontal, or left/right translations are formed by precomposing fwith g , or f (x + c).If c is positive, then the graph of f will be shifted to the left.If c is negative, then the graph of f will be shifted to the right.Let’s look at an example.

Horizontal Translations, Example 1 (x2)

This is the graph of f (x) = x2

The graph of h(x) = (x + 2)2 will be shifted left two units.



The graph of h(x) = (x + 2)2 will be shifted left two units.

h(−2) = f (0)



The graph of h(x) = (x + 2)2 will be shifted left two units.h(−2) = f (0)



The graph of j(x) = (x − 3)2 will be shifted right three units.



The graph of j(x) = (x − 3)2 will be shifted right three units.

j(3) = f (0)



The graph of j(x) = (x − 3)2 will be shifted right three units.j(3) = f (0)

Horizontal Translations, Example 2

This is the graph of some function f

The graph of h(x) = f (x + 2) will be shifted left two units.



The graph of h(x) = f (x + 2) will be shifted left two units.

h(−2) = f (0)



The graph of h(x) = f (x + 2) will be shifted left two units.h(−2) = f (0)



The graph of j(x) = f (x − 2) will be shifted right two units.



The graph of j(x) = f (x − 2) will be shifted right two units.

j(2) = f (0)



The graph of j(x) = f (x − 2) will be shifted right two units.j(2) = f (0)

Vertical Translations

We saw that f (x + c) is a horizontal translation.

We will see that f (x) + c will be a vertical translation.If c is positive, then the graph of f will be shifted up.If c is negative, then the graph of f will be shifted down.Let’s look at an example.


We saw that f (x + c) is a horizontal translation.We will see that f (x) + c will be a vertical translation.

If c is positive, then the graph of f will be shifted up.If c is negative, then the graph of f will be shifted down.Let’s look at an example.


We saw that f (x + c) is a horizontal translation.We will see that f (x) + c will be a vertical translation.If c is positive, then the graph of f will be shifted up.

If c is negative, then the graph of f will be shifted down.Let’s look at an example.


We saw that f (x + c) is a horizontal translation.We will see that f (x) + c will be a vertical translation.If c is positive, then the graph of f will be shifted up.If c is negative, then the graph of f will be shifted down.



We saw that f (x + c) is a horizontal translation.We will see that f (x) + c will be a vertical translation.If c is positive, then the graph of f will be shifted up.If c is negative, then the graph of f will be shifted down.Let’s look at an example.

Vertical Translations, Example 1 (x2)

Again we have the graph of f (x) = x2.

The graph of h(x) = x2 + 3 will be shifted up three units.

Vertical Translations, Example 1 (x2)

Again we have the graph of f (x) = x2.

The graph of j(x) = x2 − 3 will be shifted down three units.

Vertical Translations, Example 2

Again we have the graph of some function f .

The graph of h(x) = f (x) + 2 will be shifted up two units.

Vertical Translations, Example 2

Again we have the graph of some function f .

The graph of j(x) = f (x)− 1 will be shifted down one unit.

Reflections

Multiplying by −1 will make negative numbers positive andpositive numbers negative.

Thus, if we compose f (x) with the function g(x) = −x we will”flip” or reflect about the x-axis or y -axis.g ◦ f (x) = −f (x) means the points in the range of f are flipped,and we reflect about the x-axis.f ◦ g(x) = f (−x) means the points in the domain of f are flipped.We would reflect about the y -axis.Let’s look at an example.

Reflections

Multiplying by −1 will make negative numbers positive andpositive numbers negative.Thus, if we compose f (x) with the function g(x) = −x we will”flip” or reflect about the x-axis or y -axis.

g ◦ f (x) = −f (x) means the points in the range of f are flipped,and we reflect about the x-axis.f ◦ g(x) = f (−x) means the points in the domain of f are flipped.We would reflect about the y -axis.Let’s look at an example.

Reflections

Multiplying by −1 will make negative numbers positive andpositive numbers negative.Thus, if we compose f (x) with the function g(x) = −x we will”flip” or reflect about the x-axis or y -axis.g ◦ f (x) = −f (x) means the points in the range of f are flipped,and we reflect about the x-axis.

f ◦ g(x) = f (−x) means the points in the domain of f are flipped.We would reflect about the y -axis.Let’s look at an example.

Reflections

Multiplying by −1 will make negative numbers positive andpositive numbers negative.Thus, if we compose f (x) with the function g(x) = −x we will”flip” or reflect about the x-axis or y -axis.g ◦ f (x) = −f (x) means the points in the range of f are flipped,and we reflect about the x-axis.f ◦ g(x) = f (−x) means the points in the domain of f are flipped.We would reflect about the y -axis.


Reflections

Multiplying by −1 will make negative numbers positive andpositive numbers negative.Thus, if we compose f (x) with the function g(x) = −x we will”flip” or reflect about the x-axis or y -axis.g ◦ f (x) = −f (x) means the points in the range of f are flipped,and we reflect about the x-axis.f ◦ g(x) = f (−x) means the points in the domain of f are flipped.We would reflect about the y -axis.Let’s look at an example.

Reflections, Example 1 (x2)

Let’s look at the graph of f (x) = x2 a third time.

h(x) = −(x2) sends the positive y -values to negative y -values andwe reflect about the x-axis.



j(x) = (−x)2 sends the positive x-values to negative x-values andwe reflect about the y -axis.



It shouldn’t be suprising that the graphs of f (x) and f (−x) are thesame.

f (x) = x2 is called an even function.



It shouldn’t be suprising that the graphs of f (x) and f (−x) are thesame. f (x) = x2 is called an even function.

Reflections, Example 2

Let’s look at the graph of f a third time.

h(x) = −f (x) sends the positive y -values to negative y -values andwe reflect about the x-axis.



j(x) = f (−x) sends the positive x-values to negative x-values andwe reflect about the y -axis.



−f (x) = f (−x), f (x) is called an odd function.

Even and Odd Functions

Definition

A function f is even if f (x) = f (−x).

Example

x2, cos(x), cos(x) + x2 are even functions.

Definition

A function f is odd if −f (x) = f (−x).

Example

x3, sin(x), sin(x) + x5 are odd functions.

Graphing Using Translations

We are capable of graphing new functions if the transformationsare given to us, but how can we use the transformations to graphsomething like f (x) = −x2 − 2x + 8?

We can immediately see one transformationf (x) = −(x2 + 2x − 8).We can also factor to see that f (x) = −(x − 2)(x + 4), but it ismore helpful if we write f (x) = −(x + b)2 + c in order easily applythe transformations.Written as above we can see we will translate horizontally by b,reflect about the x-axis, then translate vertically by c .The process we will use is called completing the square.


We are capable of graphing new functions if the transformationsare given to us, but how can we use the transformations to graphsomething like f (x) = −x2 − 2x + 8?We can immediately see one transformationf (x) = −(x2 + 2x − 8).

We can also factor to see that f (x) = −(x − 2)(x + 4), but it ismore helpful if we write f (x) = −(x + b)2 + c in order easily applythe transformations.Written as above we can see we will translate horizontally by b,reflect about the x-axis, then translate vertically by c .The process we will use is called completing the square.


We are capable of graphing new functions if the transformationsare given to us, but how can we use the transformations to graphsomething like f (x) = −x2 − 2x + 8?We can immediately see one transformationf (x) = −(x2 + 2x − 8).We can also factor to see that f (x) = −(x − 2)(x + 4), but it ismore helpful if we write f (x) = −(x + b)2 + c in order easily applythe transformations.

Written as above we can see we will translate horizontally by b,reflect about the x-axis, then translate vertically by c .The process we will use is called completing the square.


We are capable of graphing new functions if the transformationsare given to us, but how can we use the transformations to graphsomething like f (x) = −x2 − 2x + 8?We can immediately see one transformationf (x) = −(x2 + 2x − 8).We can also factor to see that f (x) = −(x − 2)(x + 4), but it ismore helpful if we write f (x) = −(x + b)2 + c in order easily applythe transformations.Written as above we can see we will translate horizontally by b,reflect about the x-axis, then translate vertically by c .

The process we will use is called completing the square.


We are capable of graphing new functions if the transformationsare given to us, but how can we use the transformations to graphsomething like f (x) = −x2 − 2x + 8?We can immediately see one transformationf (x) = −(x2 + 2x − 8).We can also factor to see that f (x) = −(x − 2)(x + 4), but it ismore helpful if we write f (x) = −(x + b)2 + c in order easily applythe transformations.Written as above we can see we will translate horizontally by b,reflect about the x-axis, then translate vertically by c .The process we will use is called completing the square.

Completing the Square, Example 1 (−x2 − 2x + 8)

First make the coefficient in front of x2 is 1.

−x2 − 2x + 8 = −(x2 + 2x − 8)Second, we find b by dividing the coefficient in front of x by 2, sob = 1 and then add 0 by adding and subtract b2.−(x2 + 2x − 8) = −(x2 + 2x + 1− 1− 8)Third, factor our perfect square and simplify to find c .−(x2 + 2x − 8) = −(x2 + 2x + 1− 1− 8) = −((x + 1)2 − 9) =−(x + 1)2 + 9To graph we start with x2, shift left 1, reflect down about thex-axis, then shift up 9.


First make the coefficient in front of x2 is 1.−x2 − 2x + 8 = −(x2 + 2x − 8)

Second, we find b by dividing the coefficient in front of x by 2, sob = 1 and then add 0 by adding and subtract b2.−(x2 + 2x − 8) = −(x2 + 2x + 1− 1− 8)Third, factor our perfect square and simplify to find c .−(x2 + 2x − 8) = −(x2 + 2x + 1− 1− 8) = −((x + 1)2 − 9) =−(x + 1)2 + 9To graph we start with x2, shift left 1, reflect down about thex-axis, then shift up 9.


First make the coefficient in front of x2 is 1.−x2 − 2x + 8 = −(x2 + 2x − 8)Second, we find b by dividing the coefficient in front of x by 2, sob = 1 and then add 0 by adding and subtract b2.

−(x2 + 2x − 8) = −(x2 + 2x + 1− 1− 8)Third, factor our perfect square and simplify to find c .−(x2 + 2x − 8) = −(x2 + 2x + 1− 1− 8) = −((x + 1)2 − 9) =−(x + 1)2 + 9To graph we start with x2, shift left 1, reflect down about thex-axis, then shift up 9.


First make the coefficient in front of x2 is 1.−x2 − 2x + 8 = −(x2 + 2x − 8)Second, we find b by dividing the coefficient in front of x by 2, sob = 1 and then add 0 by adding and subtract b2.−(x2 + 2x − 8) = −(x2 + 2x + 1− 1− 8)

Third, factor our perfect square and simplify to find c .−(x2 + 2x − 8) = −(x2 + 2x + 1− 1− 8) = −((x + 1)2 − 9) =−(x + 1)2 + 9To graph we start with x2, shift left 1, reflect down about thex-axis, then shift up 9.


First make the coefficient in front of x2 is 1.−x2 − 2x + 8 = −(x2 + 2x − 8)Second, we find b by dividing the coefficient in front of x by 2, sob = 1 and then add 0 by adding and subtract b2.−(x2 + 2x − 8) = −(x2 + 2x + 1− 1− 8)Third, factor our perfect square and simplify to find c .

−(x2 + 2x − 8) = −(x2 + 2x + 1− 1− 8) = −((x + 1)2 − 9) =−(x + 1)2 + 9To graph we start with x2, shift left 1, reflect down about thex-axis, then shift up 9.


First make the coefficient in front of x2 is 1.−x2 − 2x + 8 = −(x2 + 2x − 8)Second, we find b by dividing the coefficient in front of x by 2, sob = 1 and then add 0 by adding and subtract b2.−(x2 + 2x − 8) = −(x2 + 2x + 1− 1− 8)Third, factor our perfect square and simplify to find c .−(x2 + 2x − 8) = −(x2 + 2x + 1− 1− 8) = −((x + 1)2 − 9) =−(x + 1)2 + 9

To graph we start with x2, shift left 1, reflect down about thex-axis, then shift up 9.


First make the coefficient in front of x2 is 1.−x2 − 2x + 8 = −(x2 + 2x − 8)Second, we find b by dividing the coefficient in front of x by 2, sob = 1 and then add 0 by adding and subtract b2.−(x2 + 2x − 8) = −(x2 + 2x + 1− 1− 8)Third, factor our perfect square and simplify to find c .−(x2 + 2x − 8) = −(x2 + 2x + 1− 1− 8) = −((x + 1)2 − 9) =−(x + 1)2 + 9To graph we start with x2, shift left 1, reflect down about thex-axis, then shift up 9.

Completing the Square, Example 2 (ax2 + bx + c)

Again, step 1 is factor out the coefficient a

ax2 + bx + c = a(x2 + ba + c

a )

Find the horizontal shift by dividing ba by 2 and keep equality by

adding 0 cleverly.

a(x2 + ba + c

a ) = a(x + ba + b

2a

2 − b2a

2+ c

a ).Factor and simplify

a(x + ba + b

2a

2− b2a

2+ c

a ) = a((x + b2a)2+ c

a−b2a

2) = a(x + b

2a)2+c− b2

4a .


Again, step 1 is factor out the coefficient aax2 + bx + c = a(x2 + b

a + ca )


adding 0 cleverly.

a(x2 + ba + c

a ) = a(x + ba + b

2a

2 − b2a

2+ c


a(x + ba + b

2a

2− b2a

2+ c

a ) = a((x + b2a)2+ c

a−b2a

2) = a(x + b

2a)2+c− b2

4a .



a + ca )


adding 0 cleverly.

a(x2 + ba + c

a ) = a(x + ba + b

2a

2 − b2a

2+ c

a ).

Factor and simplify

a(x + ba + b

2a

2− b2a

2+ c

a ) = a((x + b2a)2+ c

a−b2a

2) = a(x + b

2a)2+c− b2

4a .



a + ca )


adding 0 cleverly.

a(x2 + ba + c

a ) = a(x + ba + b

2a

2 − b2a

2+ c


a(x + ba + b

2a

2− b2a

2+ c

a ) = a((x + b2a)2+ c

a−b2a

2) = a(x + b

2a)2+c− b2

4a .

Exponential Functions

Definition

Let b > 0, an exponential function is a function of the form

f (x) = bx

where b is the base, and x is the exponent.

Exponential functions with b > 1 grow incredibly fast, faster thanany polynomial.

There is a nice legend about the payment ofwheat and a chessboard that illustrates this quite nicely.

Question

What is the one point that will agree on all graphs of exponentialfunctions bx?

Question

What can you say about exponentional functions with b < 1?

http://en.wikipedia.org/wiki/Wheat_and_chessboard_problem


Exponential Functions

Definition

Let b > 0, an exponential function is a function of the form

f (x) = bx

where b is the base, and x is the exponent.

Exponential functions with b > 1 grow incredibly fast, faster thanany polynomial. There is a nice legend about the payment ofwheat and a chessboard that illustrates this quite nicely.

Question

What is the one point that will agree on all graphs of exponentialfunctions bx?

Question

What can you say about exponentional functions with b < 1?



Exponential Functions, a Question

Question

Let f be a continuous function such that f (x + y) = f (x)f (y), andf (1) = 2.

(a) What are f (2) and f (3)? If n is an integer, what is f (n)?(b) For a rational number m

n , what is f (mn )? We will see inthe section on continuity that it is ”enough” to uniquely define acontinuous function on the rationals.

Laws of Exponents

Theorem

For a, b, x , y ∈ R, a, b > 0

ax+y = axay

ax−y = ax

ay

(ax)y = axy

(ab)x = axbx

Question

Show that the second and thirds laws follow from the first.

Graphing Exponentials

Question

Which graph is the graph of the exponential with the largest base?(for ax , a is called the base)

Graphing Exponentials

Exponentials functions are affected by transformations just like anyother function.

Question

How can we write an exponential with base a < 1 as a compositionof an exponential with base b > 1 and a transformation?

Let’s look at an example

Graphing Exponentials, Example 1 f (x) =(

32

)x

Let’s look at the base graph


32

)x


The two most important points are (0, f (0)) and (1, f (1)), thoseseems right for the sketch.


32

)x


We can shift it vertically by 1.


32

)x


We can shift it vertically by 1.

Giving usg(x) = f (x) + 1 =

(32

)x+ 1.


32

)x


We can shift it vertically by 1. Giving usg(x) = f (x) + 1 =

(32

)x+ 1.


32

)x


We can shift it by 1 and horizontally by 1.


32

)x


We can shift it by 1 and horizontally by 1.

Giving us

h(x) = f (x − 1) + 1 =(32

)(x−1)+ 1.


32

)x


We can shift it by 1 and horizontally by 1. Giving us

h(x) = f (x − 1) + 1 =(32

)(x−1)+ 1.


32

)x


We can also reflect it about the y -axis.


32

)x


We can also reflect it about the y -axis.

Giving us g(x) = f (−x) =(32

)−x=(23

)x, which should answer

the preceding question.


32

)x


We can also reflect it about the y -axis.Giving us g(x) = f (−x) =

(32

)−x=(23

)x, which should answer

the preceding question.

Identifying Exponentials Algebraically

Example

Find constants C , a > 0 such that g(x) = Cax such thatg(1) = 3

√3, g(2) = 9, g(4) = 27.

√3g(1) = g(2)√

3Ca = Ca2√

3 = a

C (√

3)2 = 9

C = 3


Example


√3, g(2) = 9, g(4) = 27.

√3g(1) = g(2)

√3Ca = Ca2√

3 = a

C (√

3)2 = 9

C = 3


Example


√3, g(2) = 9, g(4) = 27.

√3g(1) = g(2)√

3Ca = Ca2

√3 = a

C (√

3)2 = 9

C = 3


Example


√3, g(2) = 9, g(4) = 27.

√3g(1) = g(2)√

3Ca = Ca2√

3 = a

C (√

3)2 = 9

C = 3

Logarithms, the Inverse of Exponentials

The title says that logarithms are inverses of exponential functions,but what does that mean?

Definition

We will define the logarithm with base a as the function loga(x)that satisfies

loga (ax) = x

aloga x = x

Question

What is the domain for loga (ax)? aloga x? (Where are theseidentities valid?)

Logarithms, the Inverse of Exponentials

The definition gives us some very important properties oflogarithms and their relations to exponentials.

Theorem

loga x = y ⇔ ay = x

loga(xy) = loga(x) + loga(y)

loga

(xy

)= loga(x)− loga(y)

loga (x r ) = r loga(x)

loga(x) = logb(x)logb(a)

Question

Use the definition and properties of exponents to prove the LogLaws.

Using Logarithms to Evaluate

Example

Solve logx 81− logx 27 = 1 for x.

1 = logx 81− logx 27 = logx81

271 = logx 3

⇒x1 = 3

x = 3


Example



27

1 = logx 3

⇒x1 = 3

x = 3


Example



271 = logx 3

⇒x1 = 3

x = 3


We will see during the section on derivatives that rewritingexponentials as powers of e is very useful.The key to rewriting exponentials as powers of e is in the naturallogarithm ln = loge

Example

Rewrite 2x as ekx

Recall that e ln x = x , so we can rewrite 2 as e ln 2.Now substitute, 2x = (e ln 2)x = ex ln 2



Example

Rewrite 2x as ekx

Recall that e ln x = x , so we can rewrite 2 as e ln 2.

Now substitute, 2x = (e ln 2)x = ex ln 2



Example

Rewrite 2x as ekx

Recall that e ln x = x , so we can rewrite 2 as e ln 2.Now substitute, 2x = (e ln 2)x = ex ln 2

Inverses

We have seen two examples of inverses already.

f (x) = bx , f −1(y) = logb y and g(x) = x2, g−1(y) =√

yWe have also seen thatf (f −1(x)) = x = f −1(f (x)) andg(g−1(x)) = x = g−1(g(x))where the compositions are definedWe can generalize the Log Law logb x = y ⇔ by = x as

Theorem

f (x) = y ⇔ f −1(y) = x .

Inverses

We have seen two examples of inverses already.f (x) = bx , f −1(y) = logb y and g(x) = x2, g−1(y) =

√y

We have also seen thatf (f −1(x)) = x = f −1(f (x)) andg(g−1(x)) = x = g−1(g(x))where the compositions are definedWe can generalize the Log Law logb x = y ⇔ by = x as

Theorem

f (x) = y ⇔ f −1(y) = x .

Inverses


√y

We have also seen thatf (f −1(x)) = x = f −1(f (x)) andg(g−1(x)) = x = g−1(g(x))where the compositions are defined

We can generalize the Log Law logb x = y ⇔ by = x as

Theorem

f (x) = y ⇔ f −1(y) = x .

Inverses


√y

We have also seen thatf (f −1(x)) = x = f −1(f (x)) andg(g−1(x)) = x = g−1(g(x))where the compositions are definedWe can generalize the Log Law logb x = y ⇔ by = x as

Theorem

f (x) = y ⇔ f −1(y) = x .

QuestionsBasic Skills

Question

Find the domain of x√x+1

Question

Let f (x) = 3x + 2, find d so that f (x + 4) = f (x) + d. What doesthis tell you about translations of linear functions.

QuestionsUnderstanding

Question

Find a function whose domain is bounded, contained in an interval(a, b), but whose range is unbounded.

Question

Find a function whose domain is bounded, but whose range is allof R.

Question

We have seen that we can translate and reflect graphs. Can werotate them? Why or Why not?

Question

If the domain of f (x) is (−4, 2), then what is the domain off (x + c)?

QuestionsUnderstanding

Question

If f (x) is an odd function, then what must f (0) equal?

Question

Why do we write a general exponentional f (x) = Cax instead off (x) = Capx?

Preliminaries About Functions

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