Intersection

[vvaann]

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

Answer key:

SPOILER: D

[reactor]

I cannot provide a full answer but I scan ay that it is either B or C (I say more probably C)
The two functions have at least one common point: y=x^12 is a parabola with the ussual shape and y=2^x intersects y-axis at (0,1). (Note that y=2^x goes to zero when x goes to minus infinity and y goes to infinity when x goes to infinity) For the y=2^x to interesct the y-axis it has first to intersect with the parabola.
For positive x's y=2^x may or may not intersect the parabola since the parabola will increase "too rapidly for x" and therefore we can say that it is restricted by two vertical lines. In any case the two functions will either have one or two at most common points. I hope this helps.

[Dragonfinity]

Easy. Just set x^12 = x^2 then
x^12 - x^2 = 0
x^2*(x^10 - 1) = 0
x^2*(x^5 - 1)*(x^5 + 1) = 0
x = 0, 1, -1
So, we have (0,0), (1,1), and (-1,1).

[vvaann]

gstergia: I also thought that there would be at most 2 common points between them. However, the offical answer says that there are 3, actually.

Dragonfinity: it's 2^x, not x^2.

[Dragonfinity]

You are right. I've done this problem before, and I remembered it being a lot more complicated. The answer is definitely 3, but it takes a little work to get to. I just did a long answer for the prime number problem, so I think I'll sit this one out. Its mostly algebra, logs, etc.

[vvaann]

Dragonfinity, thank you for the solution to the prime-number problem.

Regarding the intersection question, I also think I'll forget it at the moment as it's rather complicated. I'm interested only in the questions whose solution can be derived directly in about 2 minutes (approximated time for one question in GRE test).

Br,
vvaann

[Dragonfinity]

well, i'm sure some people can do this in 2 minutes, but it took me a bit longer!

[awhig]

Answer should be 'C' . If you draw the curves , they intersect at two points.

[matroid]

Originally Posted by awhig

Answer should be 'C' . If you draw the curves , they intersect at two points.

No! (Draw more carefully!) It's pretty obvious, that the number of intersection points must be odd, because "in minus infinity" x^12 is greater but "in plus infinity" 2^x is greater. (Certainly this is not a proof, the curves could be tangent to each other, for instance, I'm just trying to give you a hint.)

The solution is three (D) indeed, and it takes no time or exact calculation to see this. I think you missed the third intersection point (I mean the one with the largest x-coordinate.)

Cheers, Md.

[awhig]

How can you be so sure that number of intersections is odd?
Are you considering the function: f(x) = x ^12 - 2 ^x , and proving the function to be even or odd?

While drawing curves, I considered following conditions:-
a) x > 0 [One point of intesection ]
b) 0 < x < 1 [ I did not find any]
c) x < 0 [One point of intersection]
d) x = 0 [ No intersection possible as f(x) <> 0 ]