vvaann Posted February 17, 2005 Share Posted February 17, 2005 At how many points in the xy-plane do the graphs of y = x^12 and y = 2^x intersect? A. None B. One C. Two D. Three E. Four Answer key: D Quote Link to comment Share on other sites More sharing options...
reactor Posted February 17, 2005 Share Posted February 17, 2005 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: Quote Link to comment Share on other sites More sharing options...
Dragonfinity Posted February 17, 2005 Share Posted February 17, 2005 Easy. Just set x^12 = x^2 then x^12 - x^2 = 0 x^2*(x^10 - 1) = 0 x^2*(x^5 - 1)*(x^5 + 1) = 0 x = 0, 1, -1 So, we have (0,0), (1,1), and (-1,1). Quote Link to comment Share on other sites More sharing options...
vvaann Posted February 17, 2005 Author Share Posted February 17, 2005 gstergia: I also thought that there would be at most 2 common points between them. However, the offical answer says that there are 3, actually. Dragonfinity: it's 2^x, not x^2. Quote Link to comment Share on other sites More sharing options...
Dragonfinity Posted February 17, 2005 Share Posted February 17, 2005 You are right. I've done this problem before, and I remembered it being a lot more complicated. The answer is definitely 3, but it takes a little work to get to. I just did a long answer for the prime number problem, so I think I'll sit this one out. Its mostly algebra, logs, etc. Quote Link to comment Share on other sites More sharing options...
vvaann Posted February 17, 2005 Author Share Posted February 17, 2005 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 Quote Link to comment Share on other sites More sharing options...
Dragonfinity Posted February 17, 2005 Share Posted February 17, 2005 well, i'm sure some people can do this in 2 minutes, but it took me a bit longer! Quote Link to comment Share on other sites More sharing options...
awhig Posted February 17, 2005 Share Posted February 17, 2005 Answer should be 'C' . If you draw the curves , they intersect at two points. Quote Link to comment Share on other sites More sharing options...
matroid Posted February 17, 2005 Share Posted February 17, 2005 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. Quote Link to comment Share on other sites More sharing options...
awhig Posted February 17, 2005 Share Posted February 17, 2005 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 ] Quote Link to comment Share on other sites More sharing options...
Dragonfinity Posted February 17, 2005 Share Posted February 17, 2005 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! :-) Quote Link to comment Share on other sites More sharing options...
awhig Posted February 17, 2005 Share Posted February 17, 2005 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 Did not go that far. Such equation can be solved using differential equation. Quote Link to comment Share on other sites More sharing options...
Dragonfinity Posted February 17, 2005 Share Posted February 17, 2005 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. Quote Link to comment Share on other sites More sharing options...
awhig Posted February 17, 2005 Share Posted February 17, 2005 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? Quote Link to comment Share on other sites More sharing options...
awhig Posted February 17, 2005 Share Posted February 17, 2005 The above, equation is transcedental euation. It can be solved either using inerpolations or differential equation. Through differential equation, it will have a trivial solution and a general solution. Quote Link to comment Share on other sites More sharing options...
Dragonfinity Posted February 17, 2005 Share Posted February 17, 2005 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? Quote Link to comment Share on other sites More sharing options...
awhig Posted February 17, 2005 Share Posted February 17, 2005 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. Quote Link to comment Share on other sites More sharing options...
Dragonfinity Posted February 17, 2005 Share Posted February 17, 2005 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. Quote Link to comment Share on other sites More sharing options...
bluefly Posted February 23, 2005 Share Posted February 23, 2005 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 Quote Link to comment Share on other sites More sharing options...
juphuy Posted February 23, 2005 Share Posted February 23, 2005 yeah, bluefly and dragonfinity already explained it clearly. for the second intersection on the right of y-axis, it is between 74 and 75. Isn't it? :) Using continuous function and f(a)*f(b) cheers Quote Link to comment Share on other sites More sharing options...
TokaBoy Posted March 28, 2005 Share Posted March 28, 2005 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... Quote Link to comment Share on other sites More sharing options...
msafri Posted May 11, 2007 Share Posted May 11, 2007 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:) Quote Link to comment Share on other sites More sharing options...
oldmathguy Posted December 18, 2008 Share Posted December 18, 2008 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. Quote Link to comment Share on other sites More sharing options...
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