Question Statement

Find the type of the function $h(x)=2x+1$ defined on $h:\mathbb{R}\to \mathbb{R}$.

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Background and Explanation

To determine the type of a function, we check if it is one-to-one (injective) and onto (surjective). A function that satisfies both properties is called a bijective function.

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Solution

Given the function: $\begin{array}{rl}h(x)& =2x+1\\ \text{Domain of }h(x)& =\mathbb{R}\\ \text{Codomain of }h(x)& =\mathbb{R}\end{array}$

Part 1: Checking if $h$ is one-to-one (Injective)

To check if a function is one-to-one, we assume that the outputs for two different inputs are equal and see if that forces the inputs to be equal.

Let $x_1,x_2$ be in the domain such that: $\begin{array}{rl}h\left(x_1\right)& =h\left(x_2\right)\\ 2x_1+1& =2x_2+1\end{array}$

Subtracting 1 from both sides: $2x_1=2x_2$

Dividing by 2: $x_1=x_2$

Since $h(x_1)=h(x_2)$ implies $x_1=x_2$, the function $h$ is a one-to-one function.

Part 2: Checking if $h$ is onto (Surjective)

A function is onto if its range is equal to its codomain.

For the linear function $h(x)=2x+1$, as $x$ takes any real value, $h(x)$ also covers all real values. $\text{Range of }h(x)=\mathbb{R}$

Since the Range ($\mathbb{R}$) is equal to the Codomain ($\mathbb{R}$), the function is onto.

Conclusion

Because the function is both one-to-one and onto, $h(x)$ is a bijective function.

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Key Formulas or Methods Used

• One-to-one (Injective) condition: $f(x_1)=f(x_2)⟹x_1=x_2$.
• Onto (Surjective) condition: Range of $f$ = Codomain of $f$.
• Bijective function: A function that is both injective and surjective.

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Summary of Steps

1. Assume $h(x_1)=h(x_2)$ and simplify the equation to show $x_1=x_2$ (proving it is one-to-one).
2. Determine the range of the function and compare it to the given codomain (proving it is onto).
3. Identify that since it is both one-to-one and onto, it is a bijective function.