Expert: Paul Klarreich - 1/26/2007

Question
Hi!
The problem I have is this. I have access to answers, but the answers don't make sense to me. I'll work a problem out, graph it, then look at the answer, and wonder how they got theirs. This has happened several times, so I have several questions. I need you to tell me why they get their answer instead of mine. I'd love it if you could help me with all of them, but I'll understand if you can't/don't have the time. Okay?!

Now, these are all graphing-type questions.
1. h(x) = [x] over x.
I got this one mostly right, but they included a ray with their open circles connected to closed circles. I don't understand where that comes from. The ray is located on
(-1,1)and curves up slightly so that the arrow is on
(-.5, 3).

2. f(x) = x squared - [x].
I used coordinates (-3,12) (-1,2) (3,6) (0,0) for my closed circles and (-2,7) (0,1) (4,13) (1,1) for my open circles. There does not appear to be any sort of order to this graph. I don't get it.

3. G(x) = [x squared]
The coordinates I used for the closed circles are: (-2,4)
(-1,1) (0,0)  and (1,1). The coordinates I used for the open circles are: (-1,4) (0,1) (1,0) and (2,1). Their answer looks like a parabola, but without the line connecting the points (and includes open circles).

4. y = 2 to the power of 1 over x
The coordinates I used are: (1,2) (-1,1/2)
They have an open circle on (0,0) and a point on (1,3 1/2). The open circle is attached to a ray which goes up (slightly curved, as well.) and the point is in the middle of a curved line.

5. y = e to the power of 1 over x
I'm not familiar with all of the rules concerning exponential fractions, so that complicated the first part of this question and I never got around to graphing it. Can x be 0? Won't that just make the 2 a 1? If I make x 1, will it be 4?

6. y = 2 to the power of x squared
The format of the answer graph is almost identical to that of #4. The coordinates I used were: (0,1) (1,4) (2,8). They have (-1, 1/2) (-2, 1 1/3) (open circle on 0) (1,2) (1/2, 4).
 
7. F(x) = log x to the base of 10 and G(x) = x; graph F times G.
I really don't know where I'm going with this one. I found the inverse of log x to the base of 10, and used 0, and 1 as my exponents (x). But in the graph they show, they have (0,1) and (10,10).

Thank you in advance for any help offered. I really appreciate it!
katie

Answer
Questioner:   Katie
Category:  Calculus
Private:  no
 
Subject:  (PreCalculus) graphing
Question:  Hi!
The problem I have is this. I have access to answers, but the answers don't make sense to me. I'll work a problem out, graph it, then look at the answer, and wonder how they got theirs. This has happened several times, so I have several questions. I need you to tell me why they get their answer instead of mine. I'd love it if you could help me with all of them, but I'll understand if you can't/don't have the time. Okay?!

Now, these are all graphing-type questions.
1. h(x) = [x] over x.
I got this one mostly right, but they included a ray with their open circles connected to closed circles. I don't understand where that comes from. The ray is located on
(-1,1)and curves up slightly so that the arrow is on
(-.5, 3).

2. f(x) = x squared - [x].
I used coordinates (-3,12) (-1,2) (3,6) (0,0) for my closed circles and (-2,7) (0,1) (4,13) (1,1) for my open circles. There does not appear to be any sort of order to this graph. I don't get it.

3. G(x) = [x squared]
The coordinates I used for the closed circles are: (-2,4)
(-1,1) (0,0)  and (1,1). The coordinates I used for the open circles are: (-1,4) (0,1)

(1,0) and (2,1). Their answer looks like a parabola, but without the line connecting the points (and includes open circles).

4. y = 2 to the power of 1 over x
The coordinates I used are: (1,2) (-1,1/2)
They have an open circle on (0,0) and a point on (1,3/2). The open circle is attached to a ray which goes up (slightly curved, as well.) and the point is in the middle of a curved line.

5. y = e to the power of 1 over x
I'm not familiar with all of the rules concerning exponential fractions, so that complicated the first part of this question and I never got around to graphing it. Can x be 0? Won't that just make the 2 a 1? If I make x 1, will it be 4?

>> See below.

6. y = 2 to the power of x squared
The format of the answer graph is almost identical to that of #4. The coordinates I used were: (0,1) (1,4) (2,8). They have (-1, 1/2) (-2, 1 1/3) (open circle on 0) (1,2)
(1/2, 4).

7. F(x) = log x to the base of 10 and G(x) = x; graph F times G.
I really don't know where I'm going with this one. I found the inverse of log x to the base of 10, and used 0, and 1 as my exponents (x). But in the graph they show, they
have (0,1) and (10,10).

>> Yes, if  x = 10, then  y = 10 log 10 = 10(1).
>> But if x = 0, log x is undefined.


Thank you in advance for any help offered. I really appreciate it!
katie
.............................................
Hi, Katie,

For your 1-3, I am going to assume that  [x] means the 'greatest integer function',

i.e. the largest integer not greater than x.  Now the descriptions of these things are

sort of hard to make, so I'll do my best.

1. [x]/x.  In the interval  [0,1],  [x] = 0, so this is just the x-axis, or the piece

of it.  For other intervals, both [x] and x have the same sign, so all of the graph is

above the x-axis, and you get pieces of a hyperbolic-type graph which are not supposed

to be connected.  

What about those circles?  I think that if x is an integer, [x]/x = x/x = 1, unless

x=0.  So you can make your 'dots' (filled-in circles) go along  y = 1.  What if x is

not an integer, and perhaps is getting very close to but just below an integer, such

as:

f(1.99) = 1/1.99 is close to 1/2
f(2.99) = 2/2.99 is close to 1/3

So on the right side, you can start drawing the graph of y = 1/x (lightly) and put

open circles at the integers, starting at x = 2.

What if x < 0 and close to but just below an integer?

f(-0.01) = -1/-0.01 = 100.  Draw that open circle at infinity.
f(-1.01) = -2/-1.01 = 2.  Draw an open circle at  (-1.2)
f(-2.01) = -3/-2.01 = 1.5
f(-3.01) = -4/-3.01 = 1.33

Looks like a hyperbola on the left going down, for your open circles.

2.  x^2 - [x].  Now x^2 gives you a parabola, and [x] is a step function, so this is

going to look as if someone took a hatchet to the parabola and each piece gets raised

on the left side and lowered on the right side.

3. [x^2].  Now the breaks in the graph come more frequently as you go away from x=0.  

Since [...] will be a step function, you will have steps which look like a parabola

from far away.  You do not have breaks only at  x = an integer, but at every point

where  x = sqrt(an integer)

.......................
4,5 are similar.

2^(1/x) and  e^(1/x) are very much alike.


You will note that if  f(x) = e^(1/x), then  
       -1
f'(x) = --- e^(1/x)
       x^2
and f'(x) is < 0 for all x, so the graph is always falling as we move to the right.
[PRECALCULUS?  Ignore that and you can still reach the same conclusion.]

When  x -> infinity, 1/x --> 0, and you are approaching  y = 2^0  or e^0, either of

which is equal to 1.  So there is a horizontal asymptote of  y = 1.

The graph is always falling, so it approaches that from above on the right side, and

from below on the left side.

As  x-->+-0,  e^x -> e^+-infinity

But  e^+inf = infinity, and e^-inf = 0.  So there is an obvious discontinuity at

(0,0).  {How about a nice open circle there?]

That should do it.
...................................

For 2^(x^2), it's like  e^(x^2) which is your famous 'Bell curve'.  Sorry, I mean the

famous 'bell curve'.  [No capital B, it's not named after someone named Bell.]

AND IT DOES NOT LOOK LIKE NUMBER 4.

....................................
F(x) = log x to the base of 10 and G(x) = x

FG(x) =  x log x

Since the domain of  log x is x > 0, there is no graph for negative values or zero.

If x = 1,  log x = 0, so there is an x-intercept there.
IF x > 1,  log x > 0, so the graph is above the x-axis.
IF x < 1,  log x < 0, so the graph is below the x-axis.

What happens as  x --> 0?  Have you learned something called l'Hospital's rule yet?  

If so, you can easily prove that
      log x
lim   ------- = 0
x->0    1/x
[PRECALCULUS?  Ignore that and you can still reach the same conclusion.]
So the graph will approach the origin.