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Elementary FunctionsPart 1, Functions

Lecture 1.2a, Graphs of Functions: Introduction

Dr. Ken W. Smith

Sam Houston State University

Spring 2013

Smith (SHSU) Elementary Functions Spring 2013 1 / 23

Representing functions visually

If we describe our function using an equation y = f(x) with inputs x andoutputs y, then we may view the inputs, x, as elements of a horizontal linein the plane and record outputs y on a vertical line.The graph of a function in the Cartesian plane is the set of values(x, f(x)).

Combining functions with the geometry of the plane gives us a nice visualway to see and understand a function.

This idea was first introduced in the 17th century by (among others) ReneDescartes and so the plane in which we draw our graph is called theCartesian plane.


A table of values

Many graphs of functions y = f(x) can be sketched by creating a table ofvalues (x, y) and then making some reasonable assumptions as to howthese points should be connected.

For example, consider the function f(x) = x2. We can create a table ofvalues.

x f(x)

−2 4

−1 1

−0.5 0.25

0 0

0.5 0.25

1 1

2 4

We then plot the points (−2, 4), (−1, 1), (−0.5, 0.25), ... on the Cartesianplane and use these points to guide us on filling in the rest of the curve.


Graphing our quadratic

We plot the points (−2, 4), (−1, 1), (−0.5, 0.25), ... on the Cartesian planeand then fill in the rest of the curve.


Graphing more functions

Some worked examples.For each function, create a table of values (with at least 5 points, where atleast one of which does not have integer value for x) and then graph thefunction.

1 f(x) = x3 − x

2 f(x) = |x|.

3 f(x) = 3√x


Graphing a cubic

1 Graph f(x) = x3 − x ...

Solutions.

1 Here is a table of a few values for the function f(x) = x3 − x

x f(x) = x3 − x

−2 −6−1 0

−0.5 −0.6250 0

1 0

2 6


Graphing a cubic

If we connect the dots, we should get something like this:


Graphing the absolute value function

2 Graph f(x) = |x|....Here is a table of values for the function f(x) = |x|.

x f(x) = |x|−2 2

−1 1

−0.5 0.5

0 0

1 1

2 2


Graphing the absolute value function



Graphing the cube root function

3 Graph f(x) = 3√x....

Here is a table of values for the function f(x) = 3√x

x f(x) = 3√x

−8 −2−1 −1−1

8 −12

0 0

1 1

8 2


Graphing the cube root function



Finding intercepts

In the next presentation we will discuss finding intercepts of functions andlooking at places where functions rise or fall.

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Lecture 1.2b, Graphs of Functions: Finding Intercepts

Dr. Ken W. Smith


Spring 2013


Finding intercepts

An x-intercept is a place where the graph touches the x-axis, that is,where y = 0.

A y-intercept is a place where the graph intercepts the y-axis, so x = 0.

Given an equation for a function f(x), it is easy to determine they-intercept: simply compute f(0).

It can be harder to find the x-intercepts of the function since we areseeking to solve the equation f(x) = 0 for x-values.


Two examples

Here are some examples of how one might approach this problem.

Examples.Find the intercepts of the following functions

1 f(x) = x2 − 2x− 3.

2 g(x) =1

x+ 2+

2x− 3

2x+ 1+ x− 5.

1 Solutions. f(x) = x2 − 2x− 3 has y-intercept (0,−3) since

f(0) = −3.To find the x-intercepts of the function, we need to solve theequation 0 = x2 − 2x− 3.We may factor x2 − 2x− 3 = (x− 3)(x+ 1) and solve0 = (x− 3)(x+ 1).The product of two expressions is zero if and only if one of theexpressions is zero so either 0 = x− 3 or 0 = x+ 1.Thus the x-intercepts occur where x = 3 or x = −1.So the x-intercepts are (−1, 0) and (3, 0) .


A harder problem....

2 The y intercepts of the function g(x) = 1x+2 + 2x−3

2x+1 + x− 5 occurwhere x = 0 so

y =1

0 + 2+

0− 3

0 + 1+ 0− 5 =

1

2− 3− 5 = −15

2.

Thus the y-intercept is (0,−15

2).

The x-intercept is where y = 0 so we examine the equation

0 =1

x+ 2+

2x− 3

2x+ 1+ x− 5

and solve for x.Multiply both sides by the denominators x+ 2 and 2x+ 1 to cleardenominators and we have

0 = (2x+ 1) + (x+ 2)(2x− 3) + (x+ 2)(2x+ 1)(x− 5).

Expanding this out and simplifying, we have

0 = (2x+ 1) + (2x2 + x− 6) + (2x3 − 5x2 − 23x− 10)

so 0 = 2x3 − 3x2 − 20x− 15.Smith (SHSU) Elementary Functions Spring 2013 16 / 23

A harder problem....

(We continue to work on solving the equation 0 = 1x+2 + 2x−3

2x+1 + x− 5.We reached the equation

0 = 2x3 − 3x2 − 20x− 15.

Trial and error (or a graphing calculator) helps us find that x = −1 is oneof the solutions to this equation. We can then factor 2x3− 3x2− 20x− 15into (x+ 1)(2x2 − 5x− 15). (More about this in Part 2 of our course!)To solve the equation 0 = (2x2 − 5x− 15) we require the quadraticformula.

Our solutions are1

4(5± 3

√5).

So the x-intercepts (after a lot of hard work) are

(−1, 0), (5 + 3√5

4, 0) and (

5− 3√5

4, 0)

(We will look more at this problem later.)


The vertical line test

In the definition of function, we require that a function have a uniqueoutput for each input. The geometric version of this requirement is the“vertical line test”. The points in the plane corresponding to a fixedx-value form a vertical line.For example, the line x = 3 is a vertical line consisting of all points of theform (3, y) for any value of y.If we graph a function then each vertical line will touch the graph inexactly one point, at the y value y = f(x). So to see if our graph is thegraph of a function, draw vertical lines and see if these lines intersect thegraph exactly once.


The vertical line test definition of a function

In the image below a vertical line (in yellow) corresponding to a particularx-value hits the curve at exactly one point. This should be true for all theelements in the domain of the function.

(from Wikipedia, author Wybailey, available under the Creative Commons license.)


Increasing and decreasing functions

A variety of applications in science require that we understand where afunction is rising or falling and that we be able to find the local minimumsand locals maximums of a function.

We will want to know the x values for which f(x) is as large as possible(in some specific region) or where f(x) is as small as possible.

Later, in calculus, we will find a universal approach to this very importantproblem, but at this stage, we are content to draw graphs and findminimums and maximums visually.



Example. Consider the function f(x) = x4 − 8x2 with graph given below:

The graph of the quartic polynomial f(x) = x4 − 8x2.



Where is this function increasing? Where is it decreasing?What are the low points on the curve (local minimums)?What are the high points on the curve (local maximums)?We can see that the function drops to the point (−2,−16), rises to theorigin (0, 0), drops again to the point (2,−16) and then rises after that.

So the function is decreasing for x-values in the region (−∞,−2) ∪ (0, 2),and rising in the region (−2, 0) ∪ (2,∞).The local minimums are (−2,−16) and (2,−16); a local maximum occursat the point (0, 0).Smith (SHSU) Elementary Functions Spring 2013 22 / 23

Visualizing functions

In the next lesson, we will look at piecewise functions.

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Lecture 1.2c, Graphs of Functions: Piecewise Functions

Dr. Ken W. Smith


Spring 2013


Piecewise functions

Functions need not be described by an equation. Some, like the function“color” described in section 1.1, will not involve equations at all.Other functions can be described, not by a single formula, but by acollection of them.

A piecewise function is a function which is described for “pieces” of thereal line. It will obey one rule in one region and another rule in anotherregion.

One of the simplest such functions might be described by the question,“How much does it cost to mail a letter?”

Let us assume that we are mailing a letter that weights less than 4 ounces.The US Postal Service offers a price chart which is copied on the next slide.


The cost of stamps piecewise function

Figure 8. Cost of mailing a letter (December 2012.)



If the letter is between 0 and 1 ounce in weight, the cost is 45 cents.

If the letter is between 1 and 2 ounces in weight, the cost is 65 cents.

If the letter is between 2 and 3 ounces in weight, the cost is 85 cents.

If the letter is is between 3 and 3.5 ounces in weight, the cost is 105 cents.

If the letter is more than 3.5 ounces in weight (but less than 4 ounces),then...

one should use a large envelope and the cost is $1.50.



We can describe this “cost of first-class” function as follows.

f(x) =

45, if 0 ≤ x ≤ 165, if 1 < x ≤ 285, if 2 < x ≤ 3105, if 3 < x ≤ 3.5150, if 3.5 < x ≤ 4


Another piecewise function

Here is another example. Consider the function defined in “pieces” asfollows:

f(x) =

{2x+ 1, if x ≤ 0x2 + 1, if x > 0

Compute f(2), f(0), f(−2) and graph this function.

Solution.Here, if x ≤ 0 then the function obeys the rule f(x) = 2x+ 1.

Thus f(−2) = 2(−2) + 1 = −3 and f(0) = 2(0) + 1 = 1.

But if x > 0 then f(x) = x2 + 1 and so f(2) = 22 + 1 = 5.


Graphing a piecewise function

The graph of f(x) =

{2x+ 1, if x ≤ 0x2 + 1, if x > 0

appears below. Notice how

the two “pieces” are glued together; we can see the line y = 2x+ 1 in theregion where x ≤ 0 and the parabola y = x2 + 1 in the region to the rightof the y-axis.

Figure 10. A piecewise defined function.



Let’s do some more examples with piecewise functions.

Worked examples.

1 Graph the function f(x) =

{2x+ 2, if x ≤ 1

4, if x > 1

}Then compute f(0), f(1), f(2) and f(100).

Solution. First we plot some points.Since x = 0 and x = 1 both fall in the region where x ≤ 1, we have thatf(0) = 2(0) + 2 = 2 and f(1) = 2(1) + 2 = 4.

Since x = 2 and x = 100 both fall in the region where x > 1 thenf(2) = 4 and f(100) = 4.



Here is the graph of f(x) =

{2x+ 2, if x ≤ 1

4, if x > 1

}



2 Graph the function below

f(x) =

x2, if x ≤ 00, if 0 ≤ x < 11, if 1 ≤ x < 22, if 2 ≤ x < 3x, if x ≥ 3

and compute f(−1), f(0.5), f(1.5), f(2.5), f(3.5).



f(x) =


Solution.f(−1) = (−1)2 = 1 since x = −1 ≤ 0.

But f(0.5) = 0 since 0 ≤ 0.5 < 1

Similarly f(1.5) = 1, f(2.5) = 2 and f(3.5) = 3.5.

The graph is:



f(x) =


(Note the use of closed dark circle to indicate that the point (such as(1, 1)) is on the graph and the use of light, open circles to indicate that apoint (such as (1, 0) or (2, 1)) is not on the graph.)



Note the use of closed dark circles to indicate that a point (such as (1, 1))is on the graph and the use of light, open circles to indicate that a point(such as (1, 0) or (2, 1)) is not on the graph.


Visualizing functions

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