Interesting GRE problem

Question

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.

Answer 1

$x^{12} = 2^x$ (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 $2 < x < \infty$.

Answer 2

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 :-)

Answer 3

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!