Interesting GRE problem I found this in a practice GRE problem. I thought I would have a crack at it after being spoiled by the answer 

At how many points in the xy-plane do the graphs of $y = x^{12}$ and $y = 2^x$ intersect?

So I thought about doing something what most people would have done, solving for the intersection, $x^{12} = 2^x$, it became utterly hopeless. 
Then I thought about using the Intermediate Value Theorem, that is
$f=x^{12} - 2^x = 0$
I suspect for $x<0$, $x^{12} > 2^x$, so $f>0$ 
For $x=0$, $f < 0$. So by IVT, there is a root somewhere between $(-\infty,0)$
For $x>0$, $x^{12} > 2^x$, so $f>0$. So by IVT, there is another root at $(0,\infty)$
So counting, I should get 2 roots, another therefore 2 points. But the actual answer was 3. So I am inclined to believe I overlooked something very important
Note: The GRE forbids technology assistance.
 A: 
I suspect for $x<0$, $x^{12} > 2^x$, so $f>0$

Obviously false for negative $x$ close to 0.

For $x>0$, $x^{12} > 2^x$

Obviously false for large positive $x$.
But really, all you have to do is draw a graph. Go away and draw one, and then come back and tell us about it :-)
A: I thought of it this way; near the origin, $2^x$ is relatively flat, while $x^{12}$ points up sharply. This gives two points of intersection. But, eventually, exponentials outgrow any polynomial, so there must be another point of intersection where $2^x$ outgrows $x^{12}$.
Actually, I didn't think of the third root at the time either and put down 2 as my answer on the practice exam. But hey, only 12 percent of test takers get it right!
A: $x^{12} = 2^x$ (for $x$ real) is equivalent to: either $x = 2^{x/12}$ or $-x = 2^{x/12}$.  Since $2^{x/12}$ is convex, its graph intersects a straight line in $0$, $1$ or $2$ points.  
$-x$ is decreasing while $2^{x/12}$ is increasing, and $-x > 1 > 2^{x/12}$ for $x < -1$ while $-x < 2^{x/12}$ for $x > 0$, so there is exactly one real solution of $-x = 2^{x/12}$ and it is in the interval $-1 \le x \le 0$. 
$x < 0 < 2^{x/12}$ for $x < 0$, $x < 2^{x/12}$ for sufficiently large $x$, while $x > 2^{x/12}$ at $x=2$, so there are two real solutions of $x = 2^{x/12}$, one with $0 < x < 2$ and one in $2 < x < \infty$. 
A: First thing to observe is that both the functions are continuous and the graphs are "smooth" curves.  
There is one and only one point of intersection in $(- \infty, 0)$ as $x^{12}$ is increasing & $2^x$ is decreasing as  $x \rightarrow- \infty$ and $2^0 > 0^{12}$. 
Now observe that when $x \rightarrow \infty, $ both the functions increase and $2^x $ dominates $x^{12}$ eventually. Since $ 2^{12} > 2^2 $, it must be the case that the graphs intersect at some point in $(2, \infty)$. This is the only point of intersection in that interval. 
Also, $2^0 > 0^{12}$ and $2^{12} > 2^2  $ tells us that there is a point of intersection in $(0,2)$ and there is exactly one such intersection. 
So altogether there are exactly 3 points of intersection. 
