Competency-Based
Assessment Made Simple.

1. Check for One-One (Injective):

A function $f$ is one-one if $f(x_1) = f(x_2)$ implies $x_1 = x_2$ for all $x_1, x_2$ in the domain of $f$.

Let $x_1, x_2 \in \mathbb{R} - \left\{ \frac{4}{3} \right\}$ such that $f(x_1) = f(x_2)$.

Then, $$\frac{4x_1}{3x_1 + 4} = \frac{4x_2}{3x_2 + 4}$$

Cross-multiplying, we get:

$$4x_1(3x_2 + 4) = 4x_2(3x_1 + 4)$$

$$12x_1x_2 + 16x_1 = 12x_1x_2 + 16x_2$$

$$16x_1 = 16x_2$$

$$x_1 = x_2$$

Since $f(x_1) = f(x_2)$ implies $x_1 = x_2$, the function $f$ is one-one.

2. Check for Onto (Surjective):

A function $f : A \to B$ is onto if for every $y \in B$, there exists an $x \in A$ such that $f(x) = y$.

Let $y \in \mathbb{R}$ be an arbitrary element in the codomain. We want to find $x \in \mathbb{R} - \left\{ \frac{4}{3} \right\}$ such that $f(x) = y$.

$$y = \frac{4x}{3x + 4}$$

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

$$3xy + 4y = 4x$$

$$4x - 3xy = 4y$$

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

$$x = \frac{4y}{4 - 3y}$$

For $x$ to be defined, $4 - 3y \neq 0$, which means $y \neq \frac{4}{3}$.

Thus, for $y = \frac{4}{3}$, there is no $x$ in the domain such that $f(x) = y$.

Therefore, the range of $f$ is $\mathbb{R} - \left\{ \frac{4}{3} \right\}$. Since the range is not equal to the codomain $\mathbb{R}$, the function $f$ is not onto.