Is $f(x) = 2x+1$ injective? Is it surjective?

Question

How would I answer this? I know what it means to be surjective and injective. Is the function $f(x)=2x+1$ injective? Is it surjective? Give reasons for your answers. I assume they are both because there is one output for every input.

Answer 1

The answer is "It depends."

If $f:\mathbb{R}\to\mathbb{R}$ then the function is both surjective and injective. For every $x\in\mathbb{R}$ we have $$f\left(\frac{1}{2}(x-1)\right)=2\left(\frac{1}{2}(x-1)\right)+1=(x-1)+1=x.$$ Thus $f$ is surjective.

Now suppose $f(x_1)=f(x_2)$ then $$ \begin{align*} f(x_1)&=f(x_2)\\ 2x_1+1&=2x_2+1\\ 2x_1&=2x_2\\ x_1&=x_2. \end{align*} $$

Thus if the images are equal, then the preimages are equal. So $f$ is injective.

Now if $f:\mathbb{Z}\to\mathbb{Z}$, then $f$ sends all integers to odd integers. So it cannot be surjective (no even number has a preimage.)

Answer 2

I assume the domain and the range are $\mathbb R$. Injective means that for all distinct $x_1,x_2$, $f(x_1)\neq f(x_2)$. You can try proving its contraposition: if $f(x_1)=f(x_2)$ then $x_1=x_2$.

To prove that the function is surjective, you can try for every $y\in \mathbb R$, there exists an $x\in \mathbb R$ such that $f(x)=y$