# Subject: How to "analyze" a function (intersections x and y axes, minima/maxima etc..).

Date : 29 Februari, 2016
Version: 0.2
By: Albert van der Sel
Doc. Number: Note 6.
For who: for beginners.

### This note is especially for beginners. Maybe you need to pick up "some" basic "mathematics" rather quickly. So really..., my emphasis is on "rather quickly". So, I am really not sure of it, but I hope that this note can be of use. Ofcourse, I hope you like my "style" and try the note anyway.

This note: Note 6: How to "analyze" (or investigate) a function?

Each note in this series, is build "on top" of the preceding ones.
Please be sure that you are on a "level" at least equivalent to the contents up to, and including, note 5.

Quite a few elements of the former notes will come together here, and that's why it's really important
to have read the former notes first.

This note will be a very small note. Using two example functions, we will explore how a typical analysis
takes place. Here you must think on how to find the minimum and maximum values of a function, or finding the intersections
with the x- and y axis, and that sort of material.

In Europe, such an investigation of a function usually means (not a full listing):
1. Finding the intersection with the y-axis.
2. Finding the intersection with the x-axis.
3. Finding all (local) minima and maxima that a function might have. Depending on the function, there might be
0, or 1, or 2, or even more minima/maxima. For example, for a parabola, we have only one (local) minimum (or maximum),
namely, the "top" of the "nose".
4. If it exist, investigate any 'asymptotic' behaviour. For example, f(x)=sin(x) does not expose an asymptotic behaviour,
and our well-known parabola (a quadratic function) does not have it either. However, a function like f(x)=1/x does have it.
5. Identify all regions (along the x-axis) where a function exposes a 'climbing' character, and where it 'descends'.
Alright, this is not a full listing, and not all Highschools in all countries use exactly the same "view" on the matter.
However, I will stick to the above (using the above points, except 5), and let's see how it works out for two nice examples.

# 1. Example 1.

Our first example function, will be f(x)=x3 - 4x.

Below you see a plot of this function. Note that it crosses the x-axis at three points, the y-axis at one point,
and, that it has one local maximum, and one minimum. For the latter points: you see that "top" near x=-1, and
the crest near x=1? Those are the maximum and minimum we are looking for.

Figure 1. Plot of the function f(x)=x3 - 4x ### 1.1 Intersection of f(x) with the y-axis.

This one is really easy. Where f(x) intersects with the y-axis, it must hold that "x=0".
So:

y=f(x)=x3 - 4x. So, if we fill in x=0, then it follows that y=0.

Thus, for x=0 and y=0, the function intersects with the y-axis. It's the point (0,0) in the XY plane.

### 1.2 Intersection of f(x) with the x-axis.

Where f(x) intersects with the x-axis, it must hold that "y=0".
So:

x3 - 4x = 0   =>
x(x2 - 4) = 0   =>

Thus: x=0 or x2 - 4 = 0

We need to solve the second part: For x2 - 4 = 0, it follows that x2=4, thus x=2 or x=-2

We have found the following x: x = 0, x = -2, and x = 2, for which it is true that y = 0.

So, at the following points in the XY plane, f(x) intersects the x-axis: (-2,0), (0,0), and (2,0).
Check the plot in figure 1, if this is indeed correct.

### 1.3 Finding the local minima and maxima of f(x) (if they exist).

Here is where we use the theory of note 5: the derivative of a function.

In note 5, we have found that the derivative of a function describes the "gradient" (or slope), of a tangent line.
At a maximum or minimum of the function, we have exactly the condition that the gradient is '0'.
Indeed, at those points the tangent line is parallel to the x-axis, with 0 slope.

Yes, you can try to picture several tangent lines along f(x) in figure 1, but you will only find two locations
where the gradient is exactly '0'.

So, we need to determine the derivative of the function, and find where it is '0'.

f(x)=x3 - 4x   =>
f'(x)=3x2 - 4

f'(x)=0 means:
3x2 - 4 = 0   =>
3x2 = 4   =>

Thus:

x2 = 4/3   =>

x=√ 4/3 or x= -√ 4/3.

So, we have found the following x's:   x=√ 4/3 and -√ 4/3 for which a maximum or minimum exists.

Now, these answer usually suffice. However, a teacher may request that you further calculate these square roots.
No problem, then you punch it into your calculator. And with a two digit accuracy (behind the dot) we find:

x = 1.15 and x = -1.15

Check in figure 1, if this is really correct.

Ok, we found the x's for the maximum and minumum. If it is requested to specify the points in the XY plane,
then for each of these two x's, we need to calculate f(x).

Now, you may feed in x = 1.15 and x = -1.15, into the equation x3 - 4x, but since those were approximations,
I personally would feed in x=√ 4/3 and -√ 4/3 into that equation.

I don't request to calculate it. I only want that you followed the reasoning sofar.
So, as an approximation:
you might say that for x = 1.15, we find that "y" is about -3.07, and for x = -1.15, we find that "y" is about +3.07.
Now, check the plot in figure 1, if this is a "reasonable" approximation.

But which point is the maximum, and which is the minimum? True, we have not answered this one.
If you have a plot, it is easy to answer. But if you do not have a plot, you still need to be able
to determine which is the local maximum, and which is the local minimum.

However, for now, if you followed the reasoning sofar, it's good enough for me.

### 1.4 Determining if f(x) exhibits asymptotic behaviour.

If you look at figure 1, and look at some larger values of "x", you see that f(x) is climbing steeper and steeper.
But is that asymptotic behaviour? No. You can keep calculating a valid "y", for any increasing value of "x".
True, the figure might suggest, that ultimately, it goes insanely fast upwards. Yes but you still can calculate an "y"
for a larger "x".

A vertical asymptote, is when you let x go near a value of x=a, where the function then really "threatens" to go to +/- infinity.

Or, when viewing the limit:

limx ↓ a f(x) --> +/- infinity

limx ↑ a f(x) --> +/- infinity

So, if we see a "singularity", or an "undefined territory" we might have asymptotic behaviour.
A good example is the function f(x)=1/x. See figure 2 below.

Figure 2. True asymptotic behaviour of f(x) when x nears '0'. Here, if we consider "limx ↓ 0 1/x", or "limx ↑ 0 1/x", we see
that f(x) really goes to +/- infinity

### 1.5 Ranges of x, for which f(x) is positive and negative.

----INTERMEZZO:

In note 1, I introduced a few mathematical symbols. I like to repeat them again:

The examples below use, for example, '0' and '2' as some "markerpoint", but it can ofcourse be any number.
The whole purpose is to introduce the ">" (larger than), and "<" (smaller than) symbols, and a few other symbols,
which can be of great help to "label" certain ranges.

Examples:

 x > 0 means: x is larger than '0' x ≥ 0 means: x is larger or equal to '0' x < 0 means: x is smaller than '0' x ≥ 0 means: x is larger or equal to '0' x ≤ 0 means: x is smaller or equal to '0' -5 < x < 5 means: x is smaller than '5', but larger than '-5' a ∧ b means: a AND b a ∨ b means: a OR b ∀ x means: for all x (in some set) ∃ x means: there exists some x (in some set)

----END INTERMEZZO.

From figure 1, or the calculations of the intersections with the x-axis, we can identify the ranges of 'x',
where f(x) is positive (y>0), or where f(x) is negative (y<0).
If you like, refer to figure 1 again.

for x<-2, then f(x) is negative.
for -2 < x < 0, then f(x) is positive.
for 0 < x < 2, then f(x) is negative.
for x>2, then f(x) is positive.
f(x) = 0 for x=-2, x=0, x=2

# 2. Example 2.

This one is actually simpler than example 1. Here we are going to investigate a simple quadratic function.

Suppose we have the function f(x)=x2 - 9x + 18.
Let's investigate it.

### 2.1 Intersection of f(x) with the y-axis.

Then we must have x=0, thus if we fill in '0' into the equation we find y=f(x)=18.

So, when x=0, we have y=18, where f(x) intersects the y-axis. It's the point (0,18) in the XY plane.

### 2.2 Intersection of f(x) with the x-axis.

Where f(x) intersects with the x-axis, it must hold that "y=0".
Thus:

x2 - 9x + 18 = 0

Using the knowledge from note 3, we know that the solutions are:

 x1= -b + √(b2 -4ac) -------------------               2a

 x2= -b - √(b2 -4ac) -------------------               2a

You certainly may use those formula's to find both intersections.

However, I "see" that "x2 - 9x + 18" can be quickly factored in (x-3)(x-6).

(x-3)(x-6) = x2 - 6x -3x + 18 = x2 - 9x + 18

So:

x2 - 9x + 18 = 0   =>
(x-3)(x-6) = 0   =>

So, at x=3 and x=6, the function intersects with the x-axis. These are the points (3,0) and (6,0) in the XY plane.

### 2.3 Finding the local minima and maxima of f(x) (if they exist).

A parabola, has one maximum, or one minimum only. From note 3 you may remember that a quadratic equation
having "x2" has it's "nose" pointed downwards, so that's a minimum.
Also, a quadratic equation having "- x2", has it's "nose" pointed upwards, so that's a maximum.

So, with this function, we have a minimum "somewhere". To find it:

f(x)=x2 - 9x + 18, thus:
f '(x)=2x -9

f'(x)=0   =>
2x -9=0   =>

Thus we have the minimum at x=4.5.

To find the corresponding y value: f(4.5)= 4.52 - 9x4.5 + 18 = 20.25 - 40.5 + 18 = -2.25.

So, at the point (4.5, -2.25) in the XY plane, the function has it's minimum.

### 2.4 Determining if f(x) exhibits asymptotic behaviour.

This one is for you. Do you think f(x) exhibits asymptotic behaviour?