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.
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.
Now it's your turn to apply the same kind of reasoning to the other problem.