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I have to prove that $f \colon x \mapsto e^{4x} + x^5 + 2$ ($f\colon \mathbb{R}\to\mathbb{R}$) is bijective. The argument given in the solution is that since the first two summands of the image is a bijective function of $x$, then so is $f$. Nonetheless, this "proof" doesn't seem at all rigorous to me, since there are many counterexamples to this argument.

So I proved that $f$ is strictly increasing by looking at its derivative, thus injective, and that every $c \in \mathbb{R}$ has at least one preimage, applying Bolzano's theorem to $f(x) - c$ and evaluating its limit at $-\infty$ and $+\infty$, and so, demonstrating $f$ to be surjective. In consequence, $f$ is bijective.

I am much more pleased about this proof than the one given in the solutions, but I want to know if I missed something, or if my hypothesis are insufficient. I gave an example to illustrate my argument, but the question I want to ask in the general form is in the title. Also, I would like to know whether the converse holds as well.

Thanks.

Update: I've been thinking about this, and realized that only monotonicity and continuity (together with unboundedness, as Mariano pointed out) are necessary for bijectiveness, and derivability only helps to prove monotonicity. Is this correct?

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Consider the $\arctan$ function. There is one condition you can add to monotinicity+derivability which will make things work... bijections of $\mathbb R$ are unbounded both above and below. –  Mariano Suárez-Alvarez Feb 12 '11 at 16:48
    
Well detailed, +1. Wish I could help. –  fdart17 Feb 12 '11 at 16:49
    
@your update: Right. –  Soarer Feb 12 '11 at 17:26
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The word is "differentiability", isn't it? –  Rahul Feb 12 '11 at 17:28
    
@Rahul As far as single-variable calculus is concerned, I think derivability and differentiability are the same. Although I haven't taken multivariable calculus, so I wouldn't know for sure. In Spanish we say derivabilidad (derivability) and diferenciabilidad (differentiability). And here derivabilidad seems much more widely used than diferenciabilidad, at least for single-variable calculus. I wanted to stick to the vocabulary in use, but the difference between the two concepts is not something I know or need to know right now, since I think it's nonexistent in this context. –  Abel Feb 12 '11 at 18:13

1 Answer 1

up vote 6 down vote accepted

A function $f: \mathbb{R} \rightarrow \mathbb{R}$ which is strictly monotonic -- i.e., either (for all $x_1,x_2 \in \mathbb{R}$, $x_1 < x_2 \implies f(x_1) < f(x_2)$) or (for all $x_1,x_2 \in \mathbb{R}$, $x_1 < x_2 \implies f(x_1) > f(x_2)$) -- is necessarily injective. This is immediate from the definition, and no continuity or differentiability is necessary for this.

But what about bijectivity? Let me assume WLOG that $f$ is strictly increasing. Then a necessary condition for bijectivity is that $\lim_{x \rightarrow \pm \infty} f(x) = \pm \infty$. (This is not absolutely immediate, but I hope and believe it will become clear after only a little thought.) For future use, let us call this the infinite limits property.

This observation can be used to see that the answer to your title question is no: as Mariano says, consider the arctangent function.

Moreover, being strictly increasing and having the infinite limits property is not enough to guarantee surjectivity: consider for instance the function defined piecewise as $x$ for $x \leq 0$ and as $x+1$ for $x > 0$. This example seems to indicate that we are missing the condition of continuity.

Proposition: Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be strictly increasing and have the infinite limits property. The following are equivalent:
(i) $f$ is continuous.
(ii) $f$ is surjective.

Since the word "homework" is being thrown around, I leave the proof to the OP. (Anyway, it is a fun and not so difficult exercise.)

As the OP says, differentiability is not part of the fundamental picture here, but comes up in practice because having strictly positive derivative on an interval ensures that a function is strictly increasing.

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