How to show that this function gets every real value exactly once? $$f(x) = {e^{\frac{1}{x}}} - \ln x$$
I thought maybe to use the Intermediate value theorem.
Thought to create 2 functions from this one. and subtracting.
And finally I will get something in the form of 
$$f(c) - c = 0$$
$$f(c) = c$$
But how I proof that this function gets every real value but only once ?
So it must be monotonic.
What theorem should I use here ? 
Thanks.
 A: Consider the function $$f(x) = {e^{\frac{1}{x}}} - \ln(x)$$ Its derivative is $$f'(x)=-\frac{e^{\frac{1}{x}}}{x^2}-\frac{1}{x}$$ is always negative since $x>0$ (because of the logarithm); so the function always decreases. It also starts at $+\infty$ when $x\to 0$ and goes to $-\infty$ when $x\to \infty$.
You could also show that $f''(x)>0$ for any $x$.
I suppose that this is sufficient to show that you can find a single $x$ such that $f(x)=k$ for any positive or negative value of $k$.
A: $$a>b \iff \frac{1}{a} < \frac{1}{b} \iff e^{1/a} < e^{1/b} \iff e^{1/b} - e^{1/a} > 0 \iff e^{1/b} - \ln b >  e^{1/a} - \ln b$$
But $$e^{1/a} - \ln b > e^{1/a} - \ln a$$
We have the function $f(x) = \displaystyle e^{\frac{1}{x}} - \ln x$ and derivative $$f'(x)=-\frac{e^{\frac{1}{x}}}{x^2}-\frac{1}{x}$$ The derivative is always negative, since we require that $x>0$ for the function to be defined (logarithm). Note also that $$\lim_{x\to 0} f(x) = \infty$$ whilst $$\lim_{x\to \infty} f(x) = -\infty$$
Now that we've proven that the function achieves every real value (surjective) we need to prove that the function achieves every real value exactly once which is the injective property of the function. This can be done by showing that $f(a) < f(b)$ is equivalent to $a < b$.
$$\begin{align}a<b &\iff \frac{1}{a} > \frac{1}{b} \\ &\iff e^{1/a} > e^{1/b} \\ &\iff e^{1/a} - e^{1/b} > 0 \\ &\iff e^{1/a} - \ln a >  e^{1/b} - \ln a\end{align}$$
But we know that if $a<b$ then $\ln a < \ln b$ and hence $e^{1/b} - \ln a > e^{1/b} - \ln b$ so that we get $$a<b \iff e^{1/a} - \ln a > e^{1/b} - \ln a > e^{1/b} - \ln b \iff f(b) > f(a) \iff f(a) < f(b)$$
And so, since we've just shown that $$a<b \iff f(a) < f(b)$$ our function is strictly decreasing and hence injective. Proving that it achieves ever real value exactly once. 
A quick sketch of the graph verifies our result. 

A: you can use the derivative to show that the function is monotonic
the derivative is 
$ \frac{-1}{x^2} e ^{\frac{1}{x}} - \frac{1}{x} < 0$ 
so the function is monotonic
when x tends to 0 the function tends to ${\infty}$
and when x tends to ${\infty}$ the function tends to $-{\infty}$
so the function will get every real value exactly once
A: Hint: set $z=\frac{1}{x}$. 
Then $f(x) = e^{z}+\log z = g(z)$ is a continuous increasing function, since it is the sum of two continuous increasing functions. Moreover: $$\lim_{z\to 0^+}g(z)=-\infty,\qquad \lim_{z\to +\infty} g(z) = +\infty. $$
