判断函数f:ℝ→ℝ, f(x)=(1+x)/(3x-1)的单射、满射、双射性

Let’s break down each property one by one with clear algebraic reasoning:

a. Injective (One-to-One) #

To confirm injectivity, we assume $f(x_1) = f(x_2)$ and show this forces $x_1 = x_2$.

Start with the equality:
$$\frac{1+x_1}{3x_1 - 1} = \frac{1+x_2}{3x_2 - 1}$$

Cross-multiply (note: $3x_1 -1 \neq 0$ and $3x_2 -1 \neq 0$ since those $x$-values are excluded from the function’s domain):
$$(1+x_1)(3x_2 -1) = (1+x_2)(3x_1 -1)$$

Expand both sides:
$$3x_2 - 1 + 3x_1x_2 - x_1 = 3x_1 -1 + 3x_1x_2 - x_2$$

Cancel common terms ($3x_1x_2$ and $-1$) from both sides:
$$3x_2 - x_1 = 3x_1 - x_2$$

Rearrange to group like variables:
$$3x_2 + x_2 = 3x_1 + x_1$$
$$4x_2 = 4x_1$$
$$x_1 = x_2$$

Since $f(x_1) = f(x_2)$ only when $x_1 = x_2$, $f$ is injective.

b. Surjective (Onto) #

For surjectivity, we need to check if every real number $y$ can be written as $f(x)$ for some $x \in \mathbb{R}$.

Set $y = \frac{1+x}{3x-1}$ and solve for $x$:
$$y(3x -1) = 1 + x$$
$$3xy - y = 1 + x$$

Bring all $x$-terms to one side and constants to the other:
$$3xy - x = 1 + y$$
$$x(3y - 1) = 1 + y$$

Solve for $x$:
$$x = \frac{1 + y}{3y - 1}$$

Notice that when $3y -1 = 0$ (i.e., $y = \frac{1}{3}$), the denominator is zero—there’s no real $x$ that satisfies $f(x) = \frac{1}{3}$. This means $\frac{1}{3}$ is not in the range of $f$.

Since there exists a real number that isn’t mapped to by $f$, $f$ is not surjective.

c. Bijective #

A function is bijective only if it’s both injective and surjective. Since $f$ fails surjectivity, $f$ is not bijective.

内容的提问来源于stack exchange，提问作者JpDGe