Intersection

[Dragonfinity]

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 < 2^x
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 < 2^x, so there must be a second pt. of intersection.

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

[awhig]

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 < 2^x, so there must be a second pt. of intersection.

Did not go that far. Such equation can be solved using differential equation.

[Dragonfinity]

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.

[awhig]

Originally Posted by Dragonfinity

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?

[awhig]

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.

[Dragonfinity]

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 < g
for x = 2, f > g
for x = 100, f < g

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?

[awhig]

Originally Posted by Dragonfinity

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 < g
for x = 2, f > g
for x = 100, f < g

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.

[Dragonfinity]

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.

[bluefly]

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, but f(2) = 2^12 (whatever this ends up being) > 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

[juphuy]

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) < 0, the problem will be solved approximately within 2-3 mins. enough for gre time.

cheers