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
result from wolframalpha
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.
ReplyDeleteOnly 17 percent of the actual GRE test takers in 1997 got this right, worse than a random selection strategy!
ReplyDeleteProper result from wolfram:
ReplyDeletehttp://www.wolframalpha.com/input/?i=Plot[{2^x,+x^12},+{x,+-100.,+100.}]+
one vote for D
ReplyDeleteshouldn't it be Intermediate Value Theorem instead of Rolle's Theorem?
ReplyDelete---
yodogyo@gmail.com
answer is (d) ...
ReplyDeletebasic 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.
I agree with Shantanu.
ReplyDeleteX>=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.
Sorry for my grammar, english is not my native.
ReplyDeleteFirst 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.
Why are there not more than 3 intersections?
ReplyDeleteThere 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.
ReplyDeleteA similar argument shows that there is a third and last intersection point between -1 and 0, and these are all.
What is the value of 3rd point of intersection? How large is it?
ReplyDeleteThe third point of intersection occurs approximately at (74.66932553, 3.00404713 X 10^22)
ReplyDelete