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Prove that the function $$ f(x) = e^x + \ln x $$ attains every real number as its value exactly once.

First, I thought to prove that this function is a monotonic continuous function.

But then I wasn't sure if that is how to prove the result, and if it is, I wasn't sure of how exactly to prove it that way.

Daniel R
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D_R
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    It’s a function defined on $\langle0,\infty\rangle$, continuous, so having intermediate value property. And its derivative is positive, so it’s monomtone increasing. Unbounded both above and below, that does it. – Lubin Jun 18 '16 at 18:41

2 Answers2

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Clearly, $f$ is continuous and differentiable over the set of positive numbers.

$$f'(x)=e^x+\frac{1}{x}$$ If $x>0, f'(x)>0$, hence it is monotonic continuous function. Also, note that $$\lim_{x \rightarrow 0^+}f(x)=-\infty$$ and $$\lim_{x \rightarrow \infty}f(x)=\infty$$

Hence, it attains every real number exactly once.

Siong Thye Goh
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    My comment is beside the point, but I was wondering if you would tell me what it means for a function to attain every real number exactly once? Is this synonymous with an injective function? – J. Dunivin Jun 18 '16 at 20:14
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    attain every real number (surjective) and exactly once (injective). – Siong Thye Goh Jun 18 '16 at 20:23
  • Dear SiongThyeGoh.. I'm aorry fir this comment..Can you look my simple problem? Can you show me mistakes in my solution? Please, Thank you so much..!! https://math.stackexchange.com/q/2678602/456510 – MathLover Mar 06 '18 at 01:11
  • Jack's solution seems awesome isn't it? – Siong Thye Goh Mar 06 '18 at 01:20
  • I did not receive a notification because you didn't tagg my name.)..Yes Jack is high reputation and skilled user. But, I edited my question and fixed my mistakes.. – MathLover Mar 07 '18 at 23:21
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You can prove it by noting three things:

  1. The function $f \colon (0,\infty) \rightarrow \mathbb{R}$ is continuous and strictly monotonic as $$ f'(x) = e^x + \frac{1}{x} > 0 \,\,\, \forall x \in (0,\infty). $$
  2. When $x \to 0^{+}$, we have $$ \lim_{x \to 0^{+}} f(x) = -\infty. $$
  3. When $x \to \infty$ we have $$ \lim_{x \to \infty} f(x) = +\infty. $$

By using $(2)$ and $(3)$, given $y \in \mathbb{R}$ we can find $x_0 < x_1$ such that $f(x_0) < y - \frac{1}{2}$ and $f(x_1) > y + \frac{1}{2}$. By continuity on $[x_0,x_1]$, $f$ must assume all values between $f(x_0)$ and $f(x_1)$ and in particular $y$. By strict monotonicity, it will assume $y$ only once.

Without proving $(2)$ and $(3)$, the function could miss some values if it has an asymptote at infinity or at $0$. For example, $f(x) = \arctan(x)$ is monotonic on $(0,\infty)$ but its image is $(0,\frac{\pi}{2})$.

levap
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