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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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1 Answer 1

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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.

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

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It is not actually right to say that for this part so and so is increasing rapidly, and things like that. I do get your point, but an argument like that would be pretty weak unless we prove everything mathematically and not logically. –  Rohinb97 Jun 23 at 13:51
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I don't feel compelled to dot every i and cross every t in my answers here; rather, I encourage the people posting questions to take my answers as a guide, and fill in the details on their own. In any event, the phrase, "increasing rapidly" appears nowhere in my answer, so I'm not sure what you object to. There is one place where I write, "increasing more slowly", but that's followed immediately by "consider the derivative", so I don't think there's any cause for complaint about that one. I don't write answers people can copy and paste into homework; I write answers they can ponder and expand. –  Gerry Myerson Jun 23 at 23:39
    
Sorry, but i am looking for an answer which proves that there are 3 solutions. I am refering to the terms you used like "stationary" and the fact that you have supported your whole answer by saying that its increasing slow for this bit, and fast for this...and im not complaining, like i said, i do get your point, but I wanted a mathematical approach for these type of questions, as i stated in the question. I get your point. But i am interested in the approach required for these type of questions. –  Rohinb97 Jun 24 at 20:48
    
"Stationary" is a perfectly well-defined technical term, meaning the derivative is zero. I have given you a mathematical approach for this question. If there is some particular piece of what I've written that you don't understand, please ask me about it, and I'll try to expand on it. –  Gerry Myerson Jun 24 at 23:39
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Nice...I understood the point. Thanks! –  Rohinb97 Jul 8 at 12:30

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