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
You allready received nice answers focused on injectivity and surjectivity. Here a slightly different route.
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 $x\in\mathbb R$.
Surjective: For any $y\in \Bbb R$, there exists $x=\frac{4-y}{3}$ such that $f(x)=y$.
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.
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 $y$.
