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Sketching Shortcuts

Date: 09/30/2012 at 09:59:17
From: Grace
Subject: Complicated Logarithic graph!


Do you have any shortcuts for graphing this?

   f(x) = 2 - | log(subscript 2) |x + 2| |

I tried to break this up into a piecewise function, but it just became so

I really envy an upperclassman for how fast he solved this. First, he
just wrote "domain is R \ {-2} and range is (-inf, 2]." I had known this
too, and that the asymptote is x = -2. BUT HOW DID HE EXACTLY know the
graph? I mean, he drew it without even plotting points! Only later did he
draw (-1, 2) and (-3, 2) as points in his drawing.

But how? How did he know the exact shape of the curve and everything?!

Please, Doctor, I love math -- but not the graphing side of it! I am
really, really, so, so, so afraid of graphs. I think I will have a hard
time on the test.

Please help me, Dr. Math, and share with me any knowledge if you ever come
up with a "shortcut" for drawing rough sketches. 

Thanks, doctor.

Date: 09/30/2012 at 11:16:29
From: Doctor Douglas
Subject: Re: Complicated Logarithic graph!

Hi Grace,

Unless the upperclassman worked out the problem beforehand, I don't think
he knew exactly the shape of the curve immediately. But he probably worked
it systematically, using techniques and shortcuts that he developed with
experience and practice. And he did it quickly enough that it seemed
almost immediate.

I believe that you already have some experience with these techniques. For
example, if I write this, I imagine that you would immediately identify it
as a straight line:

   g(x) = 3510.0025 x + 0.0040014

Moreover, you know right away that its y-intercept is some small negative
number, and its slope is large -- some 3500 or so. You probably didn't
need to explicitly evaluate g(0), or resort to entering numbers on a
calculator. But you could generate a rough sketch of this pretty quickly.

So let's see what can be done with your function:

   f(x) = 2 - | log(subscript 2) |x + 2| |. 

The steps I go through are these:

1. What's the domain? Well, this is defined for all x, right? No, wait; if
a logarithm tries to operate on zero or a negative number, that's not
defined. So we have make sure the |x + 2| is positive. That means
excluding x = -2. But every other x is okay.

2. What's the range? Well, the outer absolute value can take on anything
between 0 and +Infinity, so f(x) must be limited to somewhere between
between 2 and -Infinity. It might not actually take on all those values in
this function (that depends on what happens with the log2 inside), but
it's a good start to limit the graph to (-Infinity, 2].

So far so good, right?  

3. Now, after a while you recognize that the effect of an expression like
|x + 2| is to make things symmetric about the vertical line at x = -2.
This makes sense; all that matters is the distance from x to -2. So we
might draw the dotted line x = -2 to indicate left-right symmetry about
this line. This is a powerful shortcut.

4. Seems like now would be a good time to plot at least some points. Let's
start with x = -2. Oh, wait: that bombs the logarithm. Where else would
this thing be easy to evaluate? We need x-values to make that logarithm
evaluate nicely. If we make the argument of the log equal to 1, it will
automatically evaluate to zero. This is why your upperclassman chose to
evaluate f at -1 and -3. Actually, he probably just chose f(-1) in order
to get log2|-1 + 2| = log2|1| = 0, and then used symmetry (see #3) 
for f(-3).

5. So far we have two points (x, y) = (-1, 2) and (-3, 2) on the graph. We
also know that this graph is symmetric about the line x = -2. How can we
draw the rest of the graph? One way is to check what happens to f(x) as 
x -> Infinity and x -> (-Infinity), as well as x -> -2:

For large positive x, we ignore the +2 in |x + 2|, and say it's
approximately x (which itself is positive), so we ignore the absolute
value signs around the x + 2, take the log, and, well, that's positive,
too. So the outer set of absolute value signs also can be ignored, and we
get something like f(x) ~= 2 - log2(x), at least for large values of x.
With practice, you will develop this skill of approximating, and of
knowing when you can ignore some quantity vs. some other quantity.

For large negative x, we can work through the absolute value signs again,
but it's easier to use symmetry from item #3 above. Symmetry can save a
lot of work, so it's important to exploit it when we have it.

And as x gets close to -2, the logarithm will evaluate to something large
and negative, and the outer set of absolute value lines will flip it to
something large and positive, so the graph of f(x) near x = -2 will look
something like 2 - [something large].

6. Okay, that is enough to at least sketch the graph, but maybe we'd like
a few more points, to confirm our thinking above, and to feel a little
better about identifying the shape of the curve. So where else will the
logarithm be easy to evaluate? Well, log2 is nice when the argument is a
power of 2, so let's make it, say, 16:

       f(14) =  2 - | log2 |14 + 2|  |
             =  2 - | log2 |16| |
             =  2 - 4
             =  -2

And we get f(-18) for very little added work as well. And maybe we should
consider a point where the log2 goes negative:

   f(-2.125) = 2 - | log2 |-2.125 + 2| |
             = 2 - | log2 |-0.125| |
             = 2 - | log2 (1/8) |
             = 2 - | (-3) |
             = 2 - 3
             = -1

You can see why I chose the value x = -2.125, right?

Selecting the right point at which to evaluate things is a very useful
skill -- you might contrive it so that something evaluates to a simple
integer; or so that something cancels out something else (often, the
numerator of some complicated fraction), greatly reducing the arithmetic
work. And now you can see why in step 4 above, I tried at first to use 
x = -2. Technically, this was an error -- zero isn't in the domain of the
logarithm -- but even the work of my mistake isn't totally lost: stuff
near log(0) should go to "large and negative."

I hope that this helps explain some of the techniques that you can use for
this problem. Please write back if you have further questions.

- Doctor Douglas, The Math Forum 
Associated Topics:
Middle School Graphing Equations
Middle School Logarithms

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