Sunday, May 23, 2010

GR9768.63: Intersections

63. 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

Solution:
We know what the graph of $2^x$ looks like: always positive, the $y$-intercept is at $(0,1)$. For $y = x^{12}$, it passes through the vertex $(0,0)$ and it concaves up. There are at most 2 intersections. Now, the two graph intersect exactly once on $(-\infty, 0)$. On the other half, $[0, \infty)$, we'll look at $f(x) = 2^x - x^{12}$ and note that $f(1) = 2^1 - 1^{12} = 1 > 0$, where as $f(2) = 2^2 - 2^{12} < 0$. Thus, by the Rolle's Theorem, we see that $f$ has a zero on $(1,2)$. Thus, $x^{12}$ intersects $2^x$ on this interval. So there are two points of intersection.
The answer is C.
-sg-
Note: the answer provided in the Answer Sheet is D, 3 points of intersection. 
result from wolframalpha

12 comments:

  1. By the same Rolle's theorem, there *is* one more point of intersection. YOu exained x=2. Now, examine x=L, where L is very large. Clearly, 2^x wins here. So, again by ROlle's theorem, there is yet another point of intersection between 2 and L. The answer is indeed, D. - Deego.

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  2. Only 17 percent of the actual GRE test takers in 1997 got this right, worse than a random selection strategy!

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  3. Proper result from wolfram:

    http://www.wolframalpha.com/input/?i=Plot[{2^x,+x^12},+{x,+-100.,+100.}]+

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  4. one vote for D

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  5. shouldn't it be Intermediate Value Theorem instead of Rolle's Theorem?

    ---

    yodogyo@gmail.com

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  6. answer is (d) ...

    basic property is that 2^x increases faster than x^12 and in fact for any a^x for which a>1 we would have a^x > x^n for any n as x--> infinity.

    Hence all we need to find is whether the graph intersects twice or once for smaller values of x. e^x and x^2 I believe intersect only once.

    2^x and x^12 intersects twice for x<2 and also for some big value of x as 2^x would be greater than x^12 for some value eventually. Hence this function intersects thrice.

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  7. I agree with Shantanu.

    X>=0: We know that 0^12< 2^0 (x=0) and 2^12>2^2 (X=2).
    Thus there is at least one intersection between 0 and 2.
    At the same time, we know that 2^x is NOT O(x^12) as x->infinity. Also, they are both positive functions, so we know that there is some x>2 such that 2^x exceeds x^12 forever.

    x<0. Its clear there is a cross in this region.

    This gives us at least 3 crosses. My gut tells me that this is all because I am familiar with the graphs of both functions. So I would definitely choose D.

    Rigorously, I do not know why there is only 3. Examining the derivatives would probably be the place to look.

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  8. Sorry for my grammar, english is not my native.
    First 2 solutions are clearly. Lets prove that there is the 3-th solution. If we prove that lim(2^x/x^12)>1 for x->+oo, we will prove that the curves intersect once more time (because if x2 is x of 2-nd intersection and x near the x2 and x>x2 than 2^x1 for x->+oo. Just use L'Hospital's rule 12 times and we have: lim(ln2^12*2^x/(x^12))=+oo for x->+oo. So, because +oo>1 than we have the 3-th point of intersection.

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  9. Why are there not more than 3 intersections?

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  10. There are precisely 3 intersections. At x=0, 2^x = 2 > 0 = x^12. At x=2, 2^x = 4 < 2^12 = x^12, hence there is one intersection point in the interval (0,2). It is a well-known fact that as x grows to infinity 2^x exceeds x^12 (quite rapidly, in fact). Hence there is a second intersection point for x>2. By drawing the graphs of these functions it is easy to see that these are all intersection points for positive values of x.

    A similar argument shows that there is a third and last intersection point between -1 and 0, and these are all.

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  11. What is the value of 3rd point of intersection? How large is it?

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  12. The third point of intersection occurs approximately at (74.66932553, 3.00404713 X 10^22)

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