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

- Each element of the domain $X$ must have one element in the range $Y$.
- Elements of the domain must not have more than one image in the range.
- Each element of the range $Y$ must have one element in the domain $X$.
- 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:

- Prove that the given function is an Injective (One-to-one) function.
- 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 $X$.

\[ f(x) = y \]

**Expert Answer:**

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:**

\[ f(x)= (x)^2+1 \]

It is a quadratic function.

**Therefore, it is not a bijection function.**