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ehrenfest
May25-08, 04:58 PM
1. The problem statement, all variables and given/known data
At how many points in the xy-plane do the graphs of y=x^12 and y=2^x intersect?


2. Relevant equations



3. The attempt at a solution
When x is positive we have x^12=e^{12 logx} and y=2^x = e^{x log 2}. When x is negative, we have x^12=e^{12 log -x} and 2^x=e^{x log 2} so we need to find all positive solutions to

x log 2 = 12 log x
-x log 2 = 12 log x

or

x = 12 log_2 x
-x = 12 log_2 x

Does anyone know how to do that...?

HallsofIvy
May25-08, 05:38 PM
Have you noticed that you are NOT asked to find the points of intersection?

What does the graph of y= x12 look like? (Or, for that matter, x to any even power.) What does the graph of y= 2x look like?

There is a very simple answer to this question that does NOT involve solving any equation.

Gokul43201
May25-08, 05:46 PM
Also, what do you know about how those functions blow up for x --> infty?

ehrenfest
May25-08, 07:23 PM
when x is 0, the first one is 0 and the second one is 1.

when x is 1, the first one is 1 and the second is 2

when x goes to minus infinity, the first one is infinity and the second is 0

when x goes to infinity, the first is infinity and the second is infinity but the second one will go there faster

So, from that we know they must cross at least once on the left half-plane. Do we know anything else? THis method might give us a lower bound, but how will it give us a proof of the answer?

rock.freak667
May25-08, 07:27 PM
as x \rightarrow + \infty
x^{12} \rightarrow ? and 2^x \rightarrow ??

When you find what "?" and "??" is, think about what that means geometrically.

ehrenfest
May25-08, 07:54 PM
Apparently you want me to say something other than infinity for you question marks and I am not really sure what.

When I draw out the graphs, I get something that looks like a parabola and an exponential. It seems like they should intersect twice "near" the origin and we want to figure out whether they intersect again on the right half-plane. Since e^{Cx} goes to infinity "faster" than any power of x, I guess they probably do. But somehow that needs to become precise.

Sorry, I really don't see what you are getting at with your question marks...

rock.freak667
May25-08, 09:21 PM
well they both tend to infinity, so that means after they intersect once on the right plane, both of the graphs just tend to infinity, meaning that they intersect only once for x>0.

ehrenfest
May25-08, 10:22 PM
OK. I'm confused. The answer is 3. What you said implies that the intersect once in right plane and twice in the left plane. I only know how to show rigorously that they intersect once in the left plane. Can you either show me how to PROVE they intersect twice in the left plane or once in the right plane?

Dick
May26-08, 12:16 AM
well they both tend to infinity, so that means after they intersect once on the right plane, both of the graphs just tend to infinity, meaning that they intersect only once for x>0.

Now that's definitely wrong.

Gokul43201
May26-08, 10:17 AM
well they both tend to infinity, so that means after they intersect once on the right plane, both of the graphs just tend to infinity, meaning that they intersect only once for x>0.This is incorrect.

The question to ask yourself (ehrenfest), is what is lim (2^x/x^12) as x---> infty?

This is really not so hard. In the right half plane there are only 3 possible values for the number of intersections, and you can eliminate one of them from the observation made at x=2. Answering the above question will eliminate the other wrong answer.

ehrenfest
May26-08, 10:42 AM
This is incorrect.

The question to ask yourself (ehrenfest), is what is lim (2^x/x^12) as x---> infty?

This is really not so hard. In the right half plane there are only 3 possible values for the number of intersections, and you can eliminate one of them from the observation made at x=2. .

That limit is infinity, so there are 2 intersections in the right half plane and one in the left. This is clear from what the graphs of x^12

So the answer is 3.

Gokul43201
May26-08, 07:43 PM
Yup.