To prove that [tex]\( f: \mathbb{R} \rightarrow \mathbb{R} \)[/tex] defined by [tex]\( f(x) = 2x - 3 \)[/tex] is a bijection, we need to show two properties: injectivity (one-to-one) and surjectivity (onto). Then, we will find the inverse function. Let's start with the proofs:
### Injectivity:
A function is injective if [tex]\( f(x_1) = f(x_2) \)[/tex] implies [tex]\( x_1 = x_2 \)[/tex].
1. Start with the assumption that [tex]\( f(x_1) = f(x_2) \)[/tex].
2. Using the definition of [tex]\( f \)[/tex], we get:
[tex]\[
2x_1 - 3 = 2x_2 - 3
\][/tex]
3. Subtract [tex]\(-3\)[/tex] from both sides:
[tex]\[
2x_1 = 2x_2
\][/tex]
4. Divide both sides by 2:
[tex]\[
x_1 = x_2
\][/tex]
Since [tex]\( x_1 = x_2 \)[/tex], the function [tex]\( f \)[/tex] is injective.
### Surjectivity:
A function is surjective if for every element [tex]\( y \)[/tex] in the codomain [tex]\( \mathbb{R} \)[/tex], there exists an element [tex]\( x \)[/tex] in the domain [tex]\( \mathbb{R} \)[/tex] such that [tex]\( f(x) = y \)[/tex].
1. Let [tex]\( y \)[/tex] be an arbitrary element in [tex]\( \mathbb{R} \)[/tex].
2. We need to find [tex]\( x \)[/tex] such that:
[tex]\[
f(x) = y \implies 2x - 3 = y
\][/tex]
3. Solve for [tex]\( x \)[/tex]:
[tex]\[
2x = y + 3 \implies x = \frac{y + 3}{2}
\][/tex]
Since [tex]\( x = \frac{y + 3}{2} \)[/tex] is a real number for any [tex]\( y \in \mathbb{R} \)[/tex], [tex]\( f \)[/tex] is surjective.
### Conclusion:
Because [tex]\( f \)[/tex] is both injective and surjective, [tex]\( f \)[/tex] is bijective.
### Finding the Inverse:
To find the inverse function [tex]\( f^{-1}(y) \)[/tex]:
1. Start with the equation for [tex]\( f(x) \)[/tex]:
[tex]\[
y = 2x - 3
\][/tex]
2. Solve for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]:
[tex]\[
2x = y + 3 \implies x = \frac{y + 3}{2}
\][/tex]
Thus, the inverse function is:
[tex]\[
f^{-1}(y) = \frac{y + 3}{2}
\][/tex]
In summary, the function [tex]\( f(x) = 2x - 3 \)[/tex] is a bijection with its inverse given by [tex]\( f^{-1}(y) = \frac{y + 3}{2} \)[/tex].
