# Determine whether each of these functions is a bijection from R to R.

1. $f(x)= −3x+4$
2. $f(x)= −3(x)^2+7$
3. $f(x)= \dfrac{x+1}{x+2}$
4. $f(x)= (x)^5 + 1$

This question aims to find which of the above-mentioned functions is a bijection from R to R.

A bijection is also known as a bijective function or one-to-one correspondence. A function is called a bijective function if it fulfills the conditions of both the “Onto” and “One-to-one” functions. For a function to be bijective, every element in the codomain must have one element in the domain such that:

$f(x) = y$

Here are some properties of the bijective function:

1. Each element of the domain $X$ must have one element in the range $Y$.
2. Elements of the domain must not have more than one image in the range.
3. Each element of the range $Y$ must have one element in the domain $X$.
4. Elements of the range must not have more than one image in the domain.

To prove that the given function is bijective, follow the steps mentioned below:

1. Prove that the given function is an Injective (One-to-one) function.
2. Prove that the given function is a Surjective (Onto) function.

A function is said to be an Injective function if each element of its domain is paired up with only one element in its range.

$f(x) = f (y)$

Such that $x = y$.

A function is said to be a Surjective function if every element of the range $Y$ has correspondence to some element in the domain $X$.

$f(x) = y$

For the given options, let’s find out which one of them is a bijective function.

Part 1:

$f(x)= −3x+4$

Firstly, let’s determine whether it is an injective function or not.

$f(y) = -3y+4$

$f(x) = f(y)$

$x = y$

Thus, it is a one-to-one function.

Now, let’s check if it is a surjective function or not.

Find out the inverse of the function:

$f(-x) = -f(x)$

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

So, it is also a surjective function.

Therefore, part 1 is a bijection function.

Part 2

$f(x)= −3(x)^2+7$

It is not a bijection function as it is a quadratic function. A quadratic function can not be a bijection.

Moreover, $f(-x) \neq -f(x)$

Therefore, part 2 is not a bijection function.

Part 3:

$f(x)= \dfrac{x+1}{x+2}$

It is also not a bijection function as there is no real number, such that:

$f(x)= \dfrac{x+1}{x+2} = 1$

Also, the given function becomes undefined when $x = -2$ as the denominator is zero. A bijective function must be defined for every element.

Therefore, part 3 is not a bijection function.

Part 4:

$f(x)= (x)^5 + 1$

It is an increasing function.

Therefore, part 4 is a bijection function.

## Example:

Determine whether each of these functions is a bijection from R to R.

$f(x)= 2x+1$

$f(x)= (x)^2+1$

For part 1:

$f(x)= 2x+1$

Let a and b \in \mathbb{R}, so:

$f(a) = f(b)$

$2a+1 = 2b+1$

$a = b$

Hence, this is an injective function.

Since the domain of this function is similar to range, therefore it is also a surjective function.

This function is a bijection function.

For part 2:

$f(x)= (x)^2+1$