Problem Overview:

We are tasked with determining whether the function $f(x) = 6x - 9$ is one-to-one (injective), onto (surjective), or both. We are also given that the domain and codomain are both the set of all real numbers.

Definitions:

• One-to-One (Injective): A function is injective if different inputs produce different outputs, i.e., if $f(x_1) = f(x_2)$ implies $x_1 = x_2$.
• Onto (Surjective): A function is surjective if every element of the codomain is the output of some element in the domain, i.e., for every $y \in \mathbb{R}$, there exists an $x \in \mathbb{R}$ such that $f(x) = y$.

Step 1: Check if $f(x) = 6x - 9$ is One-to-One (Injective)

To check if the function is injective, we assume $f(x_1) = f(x_2)$ and show that this implies $x_1 = x_2$.

$f(x_1) = f(x_2)$

$6x_1 - 9 = 6x_2 - 9$

Now, simplifying:

$6x_1 = 6x_2$

Dividing both sides by 6:

$x_1 = x_2$

Since we have shown that $f(x_1) = f(x_2)$ implies $x_1 = x_2$, the function $f(x) = 6x - 9$ is one-to-one.

Step 2: Check if $f(x) = 6x - 9$ is Onto (Surjective)

To check if the function is surjective, we need to show that for every $y \in \mathbb{R}$, there exists an $x \in \mathbb{R}$ such that $f(x) = y$.

Let $y \in \mathbb{R}$ be an arbitrary real number. We want to find $x$ such that:

$f(x) = y$

Substituting the expression for $f(x)$:

$6x - 9 = y$

Solving for $x$:

$6x = y + 9$

$x = \frac{y + 9}{6}$

Since $x = \frac{y + 9}{6}$ is a valid real number for any real $y$, it follows that for every $y \in \mathbb{R}$, there exists an $x \in \mathbb{R}$ such that $f(x) = y$. Thus, the function is onto.

Conclusion:

The function $f(x) = 6x - 9$ is both one-to-one and onto, meaning it is a bijective function.

---

Would you like any further explanations or more examples on injective and surjective functions?

Here are 5 related questions you might find helpful:

1. How can we prove a function is not one-to-one?
2. What is the difference between an injective and a surjective function?
3. Can a function be onto but not one-to-one? Provide an example.
4. What is the inverse of a bijective function, and how do we find it?
5. How can we check if a function is bijective for non-linear functions?

Tip: When checking if a function is one-to-one, try to set $f(x_1) = f(x_2)$ and solve for $x_1 = x_2$. This is often the easiest way to prove injectivity.