# What is the approach required for questions in which you least expect that the graphs meet?

Find the no. of solutions of x in these two equations:

(A)$2^x=x^2+1$

(B)$e^x=2x^2$

Both are of the same type, that is, the answer is the least you can expect. (When you plot it on a grapher, you will get it). Both are interesting scenarios but I am having a problem trying to prove it. Please give the approach required for these type of questions. And also, more examples which are even closer and more interesting will be appreciated.

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Note for $x<0$ we have $2^x<1<x^2+1$; we have equality at $x=0$, but $2^x$ is increasing here while $x^2+1$ is stationary, so $2^x>x^2+1$ for $x$ a little bigger than $0$; there's equlaity again at $x=1$, but now $2^x$ is increasing more slowly than $x^2+1$ (consider the derivative), so $2^x<x^2+1$ for $x$ a little bigger than $1$; but exponentials grow faster than polynomials, so there must be a third value of $x$, $x>1$, where the two are equal. It shouldn't be hard to show that for such $x$ and beyond, $2^x$ grows faster than $x^2+1$, so there are no more solutions than the three we have found.