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Answer the following.
If f: mathbbR → mathbbR is defined by f(x) = (2x - 7)/4, show that f(x) is one - one and onto.
If $f: \mathbb{R} \to \mathbb{R}$ is defined by $f(x) = \frac{2x - 7}{4}$, show that $f(x)$ is one - one and onto.
Given: $f(x) = \frac{2x - 7}{4}$
To show: $f(x)$ is one - one and onto
Answer
Answer
To show $f(x) = \frac{2x-7}{4}$ is one-one and onto.
- One-one: let $x_1, x_2 \in \mathbb{R}$ with $f(x_1) = f(x_2)$.
- $\frac{2x_1 - 7}{4} = \frac{2x_2 - 7}{4}$, by the definition of $f$.
- $2x_1 - 7 = 2x_2 - 7$, multiplying by 4.
- $x_1 = x_2$, so $f$ is one-one.
- Onto: let $y \in \mathbb{R}$ (codomain) and put $y = \frac{2x-7}{4}$.
- $x = \frac{4y + 7}{2}$, which is a real number for every real $y$.
- Then $f\left(\frac{4y+7}{2}\right) = \frac{(4y+7) - 7}{4} = y$, so every $y$ has a pre-image.
- Hence $f(x)$ is one-one and onto.