Let f : R → R be defined by f (x) = 1/x for all x ∈ R. What type of function is f? A) Not defined B) Surjective C) Bijective D) Injective
Understand the Problem
The question is asking to classify the function f(x) = 1/x in terms of its properties, specifically whether it is surjective, injective, bijective, or not defined.
Answer
Injective but not surjective or bijective.
Answer for screen readers
The function $f(x) = \frac{1}{x}$ is injective but not surjective or bijective.
Steps to Solve
- Define the Concepts To classify the function $f(x) = \frac{1}{x}$, we need to understand the properties of injective, surjective, and bijective functions.
- A function is injective (one-to-one) if different inputs give different outputs.
- A function is surjective (onto) if every possible output has a corresponding input.
- A function is bijective if it is both injective and surjective.
- Check Injectivity To determine if $f(x)$ is injective, assume $f(a) = f(b)$ for any $a, b$ in the domain.
$$ \frac{1}{a} = \frac{1}{b} $$
Cross-multiplying gives:
$$ b = a $$
Since $a$ must equal $b$, $f(x)$ is injective.
- Check Surjectivity Next, we need to check if $f(x)$ is surjective. The range of $f(x)$ must cover all real numbers.
Consider any real number $y$. We need to find an $x$ such that:
$$ y = \frac{1}{x} $$
This implies:
$$ x = \frac{1}{y} $$
The value of $y$ cannot be $0$, thus not every real number is an output of $f(x)$. Therefore, $f(x)$ is not surjective.
- Determine Overall Classification Since $f(x)$ is injective but not surjective, it is not bijective.
The classification of the function is as follows: $f(x) = \frac{1}{x}$ is an injective function but not surjective or bijective.
The function $f(x) = \frac{1}{x}$ is injective but not surjective or bijective.
More Information
The function $f(x) = \frac{1}{x}$ is defined for all real numbers except at $x = 0$. It uniquely maps inputs to outputs in a way that no two inputs produce the same output, but not every real number can be achieved as an output from this function.
Tips
- Assuming that the function is surjective simply because it covers many values.
- Forgetting to check the domain carefully (i.e., failing to consider that $f(x)$ is undefined at $x = 0$).
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