Jump to content
Urch Forums

Intersection


vvaann

Recommended Posts

I cannot provide a full answer but I scan ay that it is either B or C (I say more probably C)

The two functions have at least one common point: y=x^12 is a parabola with the ussual shape and y=2^x intersects y-axis at (0,1). (Note that y=2^x goes to zero when x goes to minus infinity and y goes to infinity when x goes to infinity) For the y=2^x to interesct the y-axis it has first to intersect with the parabola.

For positive x's y=2^x may or may not intersect the parabola since the parabola will increase "too rapidly for x" and therefore we can say that it is restricted by two vertical lines. In any case the two functions will either have one or two at most common points. I hope this helps. :grad:

Link to comment
Share on other sites

Dragonfinity, thank you for the solution to the prime-number problem.

 

Regarding the intersection question, I also think I'll forget it at the moment as it's rather complicated. I'm interested only in the questions whose solution can be derived directly in about 2 minutes (approximated time for one question in GRE test).

 

Br,

vvaann

Link to comment
Share on other sites

Answer should be 'C' . If you draw the curves , they intersect at two points.

 

No! (Draw more carefully!) It's pretty obvious, that the number of intersection points must be odd, because "in minus infinity" x^12 is greater but "in plus infinity" 2^x is greater. (Certainly this is not a proof, the curves could be tangent to each other, for instance, I'm just trying to give you a hint.)

 

The solution is three (D) indeed, and it takes no time or exact calculation to see this. I think you missed the third intersection point (I mean the one with the largest x-coordinate.)

 

Cheers, Md.

Link to comment
Share on other sites

How can you be so sure that number of intersections is odd?

Are you considering the function: f(x) = x ^12 - 2 ^x , and proving the function to be even or odd?

 

While drawing curves, I considered following conditions:-

a) x > 0 [One point of intesection ]

b) 0

c) x

d) x = 0 [ No intersection possible as f(x) 0 ]

Link to comment
Share on other sites

NO. The graphs of y = x ^12 and y = 2 ^x intersect twice over the positive x-axis and once over the negative x-axis. Draw them again more carefully. I am not going to take the time to give an exact solution because it is a real pain but note the following:

 

x=1: x^12 = 1, 2^x = 2, hence x^12

x=2: x^12 = 4096, 2^x = 4, hence x^12 > 2^x, so there is clearly one pt. of intersection

The following is an approximation by computer, but the missing digits are irrelevant, as you will see:

x=100: x^12 = 1E+24, 2^x = 1.27E+30, hence x^12

 

Does this end it? Or do I need to actually work out the problem! :-)

Link to comment
Share on other sites

I am making it easier. I plugged a few numbers in to help you see why the answer is 3. If you can solve it, do it. There is still 3 intersections. If you still don't get it, I will present the full solution.

 

Differential equation? Show me.

 

Plug in solutions are not always accurate. What was your approach?

Link to comment
Share on other sites

Let there be two equations on the xy axis: f(x) = x ^12 and g(x) = 2 ^x. Both are continuous and defined for all x. I showed above that

 

for x = 1, f

for x = 2, f > g

for x = 100, f

 

Since f and g are continuous, there must be 2 intersection points to the right of the y-axis. [if you don't understand why this is true, consider the function h(x) = f(x) - g(x)]. You already agreed that there is 1 intersection point to the left of the y-axis. Hence, there are 3 intersection points.

 

You are making this more complicated than it is. From your profile, I see you are not a mathematician, so why are you arguing against this so hard? If you just don't understand, say so, but I get the impression that you are telling us we are wrong. Trust me, Matroid and I are not wrong on this. We know our stuff. I am just too lazy to do the algebraic solution again, but I will take the time to work it out again if it will help you understand. Alternatively, when I get home from work tonight, I will graph it on Mathematica and post the graph if you like.

 

You're kidding about the differential equation, right? Let y = x ^12 - 2 ^x and find the zeros of the function. What does a differential equation have anything to do with this problem?

Link to comment
Share on other sites

Let there be two equations on the xy axis: f(x) = x ^12 and g(x) = 2 ^x. Both are continuous and defined for all x. I showed above that

 

for x = 1, f

for x = 2, f > g

for x = 100, f

 

Since f and g are continuous, there must be 2 intersection points to the right of the y-axis. [if you don't understand why this is true, consider the function h(x) = f(x) - g(x)]. You already agreed that there is 1 intersection point to the left of the y-axis. Hence, there are 3 intersection points.

 

You are making this more complicated than it is. From your profile, I see you are not a mathematician, so why are you arguing against this so hard? If you just don't understand, say so, but I get the impression that you are telling us we are wrong. Trust me, Matroid and I are not wrong on this. We know our stuff. I am just too lazy to do the algebraic solution again, but I will take the time to work it out again if it will help you understand. Alternatively, when I get home from work tonight, I will graph it on Mathematica and post the graph if you like.

 

You're kidding about the differential equation, right? Let y = x ^12 - 2 ^x and find the zeros of the function. What does a differential equation have anything to do with this problem?

 

It is unfortunate that you are taking me as though I am arguing.

I am just presenting my perspective. Though I do not say I am a scientist, but as far as Mathematics is concerened, I also did it . If you think I am challenging you, it is again unfortunate.

 

 

Link to comment
Share on other sites

Awhig, then what are you saying. Do you agree or disagree with the answer and our solution. This is math. There is no perspective. There are right and wrong answers, and sometimes there are unanswerable questions that cannot be proved either way. You have challenged the answers presented here, and you have made several assertions, but have presented no mathematics to back them up. So, who is arguing? Please, tell us what you don't understand, and we'll clear it up for you. If the math is too difficult for you, I am sorry.
Link to comment
Share on other sites

yeah this one can be done pretty quickly if you think about it right

 

dragonfinity has the right idea... you don't care what the points actually are and you don't want to bother taking the time to draw the graph

 

since the functions are continuous you compare the values on different intervals, like between 1 and 2 for example, f(1)=1^12 = 1 and g(1)=2^1 = 2 so f g(2) = 2^2 = 4 so there must be a point of intersection between x=1 and 2. etc. I think dragonfinity explained it more thoroughly

Link to comment
Share on other sites

  • 1 month later...

The answer is 3.

 

1. For x

- x^12 goes from +infinity to 0

- 2^x goes from 0 to 1

so the 2 curves must intersect at one point, just use the Newton theorem that helps to find zero point.

2. For 0

- for x = 0, 0^12 = 0 and 2^0=1, so 2^x is above x^12 for x = 0

- for x = 2, 2^12 = 4096 and 2^2=4, so x^12 is above 2^x for x = 2

so the 2 curves must intercept at one point between 0 and 2

3. For x>2, x-> infinity

lim 2^x > lim x^12 when x-> infinity (a simple rule says that exponential goes faster to the infinity than any x^n)

so the 2 curves must intercept at one point between 2 and the infinity.

 

Definitely the answer is D.Three(3).

 

Keep in touch...

Link to comment
Share on other sites

  • 2 years later...
I thought of it the same way as dragoninfinity/matroid (using concept of function growth order). Anyways, I am interested in the algebraic solution, so can someone please provide one (in the spirit of mathematical exactness :grad:)
Link to comment
Share on other sites

  • 1 year later...

Think roughly about the shape of the graphs. There is one intersection with

x

higher. But we know that for very large x ("in the limit") x^12 ends up lower.

So (looking at the general shape of the graphs) there must be two

intersections with x > 0. So there are three intersections altogether.

Link to comment
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Reply to this topic...

×   Pasted as rich text.   Restore formatting

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

×
×
  • Create New...