# show that $f(x)=-3x+4$ is bijective

Determine whether each of these functions is a bijection from $\mathbb{R}$ to $\mathbb{R}$

a) $f(x)=-3x+4$

So I know that a function is bijective if it is both injective (one-to-one) and surjective (onto).

A function is one-to-one if every $x$ has a unique $y$.

And it is onto if for every $y$ there is an $x$ such that $f(a)=b$.

But I don't know how write it down and show that $f(x)=-3x+4$ is bijective

• Take $y=f(x)$ and determine x in terms of y. Commented Apr 16, 2015 at 11:40
• Do you want a formal proof? Actually, a linear function $f(x) = ax+b$ is always bijective Commented Apr 16, 2015 at 11:41
• Please read the definition of injectivity: a function is one-to-one if every $y$ has at most one $x$. Commented Apr 16, 2015 at 11:43
• @AlexSilva (if $a\ne0$) Commented Apr 16, 2015 at 12:20
• @MarioCarneiro, Yes! You're right Thanks! Commented Apr 16, 2015 at 12:46

Can you find a function $g:\mathbb R\rightarrow \mathbb R$ such that $f(g(x))=x$ for each $x\in\mathbb R$?

If you have found such $g$ then check whether it is also true that $g(f(x))=x$ for each $x\in\mathbb R$.

If so then you are ready because you have shown that $f$ is "invertible" (i.e. has an inverse). A function is bijective if and only if it is invertible.

You could also take the opposite route: finding a function $g:\mathbb R\rightarrow \mathbb R$ such that $g(f(x))=x$ for each $x\in\mathbb R$ and checking whether it is also true that $f(g(x))=x$ for each $x\in\mathbb R$.

• FYI the correct term for "has an inverse" is invertible. Also your two routes involve exactly the same checks - what's the difference? Commented Apr 16, 2015 at 12:21
• @MarioCarneiro Thank you, I corrected. As you said there is no essential difference between the two routes. The OP can decide what is most easy for him: finding $g$ based on $f\circ g=1$ (and check $g\circ f=1$) or finding $g$ based on $g\circ f=1$ (and check $f\circ g=1$). Commented Apr 16, 2015 at 13:15
• @Valentino There are lots of cases in which it is quite difficult to find the inverse itself. Essential here is not finding it, but proving that it exists, which actually comes to the same as proving injectivity and surjectivity. I gave this answer because in this case it is easy to find the inverse. Secondly in my view it is good to emphasize that 'being invertible' is exactly the same thing as 'being bijective'. Commented Apr 16, 2015 at 14:32
• @ drhab I completely agree! I think things start going a lot more smoothly later on once one learns that being invertible is exactly the same thing as being bijective. That fact comes up probably every day. However, I was wondering if you have tried to prove cardinality type problems like this. If so, was it easier or harder, or just the same? Commented Apr 16, 2015 at 14:34
• @Valentino My experience in this is very 'poor'. I am just too lazy for doing things like that :) So my answer is: "no, I never tried that." I hope that it is not disappointing for you. Commented Apr 16, 2015 at 14:38
1. Surjective: For any $y\in \Bbb R$, there exists $x=\frac{4-y}{3}$ such that $f(x)=y$.

2. Injective: For any $a\not=b$, does it $-3a+4=-3b+4$ hold?

Let the function $f:\mathbb{R}\to \mathbb{R}$ be defined by $f(x)=-3x+4$.

We say that a function $f$ is injective if $\forall x_{1},x_2$ $f(x_{1})=f(x_2)$ implies $x_1 = x_2$. Hence, after some algebra we can see that $-3x_{1}+4= -3x_2+4$ implies $x_1=x_2$. So our function is injective.

Now, we say a function is surjective if for all $y\in \mathbb{R}$(range) there is some $x\in \mathbb{R}$ (domain) such that $f(x)=y$. Choose $x = \frac{4-y}{3}$, then we can see that $f(x)=y$.

A function is bijective if it is both surjective and injective. Hence, $f$ is bijective.

If you have any further questions please let me know.

• The last word of your second paragraph was supposed to be injective. Commented Apr 17, 2015 at 1:57

One-to-one:

Suppose there are two values $x_1,x_2$ such that $f(x_1)=f(x_2)$. Show that $x_1=x_2$ necessarily.

Onto:

Take $y=-3x+4$ and try to obtain $x$ in terms of $y$.

• If I try to obtain $x$ in terms of $y$ I am actually inverting it. Commented Apr 16, 2015 at 11:51