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I have no idea on how to prove that $$ x(e^x-e^{-x})-e^x=0 $$ has at least 2 real solutions in R? How would you generally go about proving solutions exist if you can't isolate the x or use the quadratic formula?

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    How about proving: $f(x) < 0$ and $f(x) \to +\infty$ as $x \to \pm \infty$. – 2017-01-12
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    how does it prove that we have solutions in R – 2017-01-12
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    It is a non-linear equation. Use the bisection method or the Newton method. – 2017-01-12
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    You also need: $f$ is continuous, and the intermediate value theorem https://en.wikipedia.org/wiki/Intermediate_value_theorem – 2017-01-12

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Multiply through by $e^x$, whihc you can do because $e^x$ is never zero for finite real $x$; so you are looking for solutions to $$ f(x) = x(e^{2x}-1) - e^{2x} = 0$$

You know that the $f(x)$ is continuous because the product of two continuous functions is continuous, as is the sum of two continuous functions, and $e^{2x}$ is continuous.

You also know that at $x=-30$, $$f(-30) = 30(1-e^{-60}) - e^{-60} > 0 $$ since $e^{-60}$ very small compared with 30. And at $x=0$, $$f(0) = - e^{0} = -1 < 0 $$ So by the intermediate value theorem, $f(x)$ is zero somewhere on the interval $(-30,0)$.

And you also know that at $x=2$ $$ f(2) = 2(e^4-1)-e^4 = e^4 -1 > 0 $$ So by the intermediate value theorem, $f(x)$ is also zero somewhere on the interval $(0,2)$.

The harder problem would be to show there are only two real zeros.

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    Ah so we use the intermediate value theorem and guess a closed bound where we know that a solution exists, and by doing that 2 times we can say that **at least** two solutions exist in R. Is that correct? – 2017-01-12
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Let's set $f(x)=x(e^x-e^{-x})-e^x$. First, look at $f(0)$: $$f(0)=0(e^0+e^{-0})-e^0=0-1=-1$$ This means that there are solutions where $f(x)<0$.

Let's take two limits: $$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}(xe^x-xe^{-x}-e^x)=\lim_{x\to\infty}(x-1)e^x-\lim_{x\to\infty}\frac{x}{e^x}=?$$ $$\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}(xe^x-xe^{-x}-e^x)=\lim_{x\to-\infty}(x-1)e^x-\lim_{x\to-\infty}\frac{x}{e^x}=?$$ If both of these are greater than $0$, then - given that $f(x)$ is continuous - there are at least two real solutions. Can you tell if this is the case?

Note that this would not guarantee that there are only two solutions, merely that at least two solutions must exist.

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    doesn't your limit to infinity diverge and wouldn't the -infinity one be undefined? – 2017-01-12
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    @SoHCahToha Yes, the limits are divergent - and you can use L'hopital's rule to calculate them. That means that there must be some points - one on either side of $x=0$ - where $f(x)=0$. – 2017-01-12
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    I don't really get why the fact that these two limit being divergent indicate that there must be some points for which f(x) = 0 – 2017-01-12
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    @SoHCahToha Both limits go to *positive* infinity. That means that as $x\to\infty$, $f(x)>0$ for large $x$. This means that $f(x)$ must have gone from negative to positive. Others have already mentioned this through the [intermediate value theorem](https://en.wikipedia.org/wiki/Intermediate_value_theorem). – 2017-01-12
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    So that tells us that the function must have crosse the x axis at least once, so how could we expand that logic to show that it crosses at least 2 times the x axis? – 2017-01-12
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    @SoHCahToha It crosses once in *each direction*. If we only had one limit, then we could only say that it crosses once. However, we have two. – 2017-01-12