## 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

#### 13 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.

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

3. Proper result from wolfram:

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

4. one vote for D

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

---

yodogyo@gmail.com

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.

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.

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.

9. Why are there not more than 3 intersections?

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.

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

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