Question

show that the relation f:R R defined by f (x)=3x+2 is a bijection also find f^-4 (x)​

Answer 1

To prove the function is surjective you must show that for any point “a” in the range there is a point “b” in the domain such that f(b)=a.

This is easily shown by letting a=3x-5 and therefore b must be (a+5)/3. Since this is a real number it is in the domain.

To prove the function is injective you must show if f(a)=c and f(b)=c then a=b.

Let f(a)=c and f(b)=c -> c=3a-5 and c=3b-5 -> 3b-5=3a-5 -> 3b=3a -> b=a

Since it is both injective and surjective we can say that f(x) is a bijection from R to R

Answer 2

$\huge \blue \gamma \pink{Answer} \blue \gamma$

First you need to show that the function is one-to-one: put into precise mathematical terms, this means

[math]f(x)=f(y)\Rightarrow x=y[/math]

Or in English: if [math]f(x)=f(y)[/math], then [math]x=y[/math]

Suppose [math]f(x)=f(y)[/math] which means

[math]3x-5-3y-5[/math] SO [math]3x=3y[/math] which means [math]x=y[/math].

Therefore, [math]f(x)[/math] is one-to-one.

Now we need to show [math]f(x)[/math] is onto [math]\mathbb{R}[/math]

This means that [math]\mathbb {R}=Rng(f)[/math]. Now the

range of f is the set of values that f maps to from the

domain. Now, we need to show that

[math]\mathbb{R}\subseteq Rng(f)[/math] (we know that

[math]Rng(f)\subseteq\mathbb {R}[/math] since f is a function)

So, suppose [math]y\in\mathbb {R}[/math] and let

[math]x=\frac{y+5){3}[/math]. Since [math]y\in\mathbb{R}[/math]

To prove the function is surjective you must show that for any point "a" in the range there is a point "b" in the domain such that f(b)=a.

This is easily shown by letting a-3x-5 and therefore b must be (a+5)/3. Since this is a real number

it is in the domain.

To prove the function is injective you must show if f(a)-c and f(b)=c then a-b.

Let f(a)=c and f(b)-c-> c=3a-5 and c-3b-5 -> 3b-5-3a-5 -> 3b-3a -> b=a

Since it is both injective and surjective we can say that f(x) is a bijection from R to R