Prove if it is a bijection.

Question

Assume that g : R → R is a bijection and define f : R → R by

f(x) = 2g(x) + 1.

Determine, with proof, whether f is a bijection.

My opinion:

I know that a function should either be increasing or decreasing in interval. So if g(x) > 0, then f(x)>0 and if g(x)<0, then f(x)<0. My thinking is that f(x) is either increasing or decreasing and hence it is also bijective.

Answer 1

I wouldn't go down the path of using derivatives for three reasons:

1. Functions may not be differentiable. It may make no sense to claim that $f'(x) > 0$ or $f'(x) < 0$.
2. Functions need not be continuous. While it is true that a continuous bijective function must be increasing or decreasing, this is not true for general functions. For example, the function $$f(x) = \begin{cases} \frac{1}{x} & \text{if } x \neq 0 \\ 0 & \text{if x = 0} \end{cases}$$ is bijective, discontinuous, and neither increasing nor decreasing.
3. Even when the functions are strictly increasing or decreasing, this can only show the functions are one-to-one. You'd have to show they are onto in a different way. For example, the function $$f(x) = \arctan(x)$$ is strictly increasing (hence one-to-one), but has a range of $(-\pi/2, \pi/2)$. Hence it is not surjective, and hence not bijective.

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Instead, simply use the definitions.

A function $f$ is one-to-one if, whenever $f(x) = f(y)$, we must have $x = y$. In our case, if we have $f(x) = f(y)$, then this means that $2g(x) + 1 = 2g(y) + 1$. By simplifying this equation, we get $g(x) = g(y)$. But, $g$ is also one-to-one, so we know the only way this can happen is if $x = y$. That is, when $f(x) = f(y)$, we have $x = y$, so $f$ is one-to-one, by definition.

A function $f$ is onto if, given any $y$, we can find an $x$ such that $f(x) = y$. So, fix $y \in \Bbb{R}$. Let's find an $x$ such that $f(x) = y$. That is, $2g(x) + 1 = y$. Solving this, we get $g(x) = \frac{y - 1}{2}$. Does there exist such an $x$? Yes, since $g$ is onto and $\frac{y - 1}{2} \in \Bbb{R}$. For this given value of $x$, rearranging the equation back again yields that $f(x) = 2g(x) + 1 = y$, which is what we want. Hence $f$ is onto.

Answer 2

If $f(x)=f(y)$ then $2g(x)+1=2g(y)+1\implies g(x)=g(y)$ and since $g$ is injective, $x=y$. Therefore, $f$ is injective.

Let $r\in\mathbb R$. Then, since $g$ is surjective, there exists $x\in\mathbb R$ such that $g(x)=\frac{r-1}{2}$. Then $f(x)=2g(x) + 1=2(\frac{r-1}{2})+1=r$ so $f$ is surjective.

Therefore, $f$ is a bijection.