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Prove the following.
Let R⁺ be the set of all positive real numbers and f: R⁺ → [4, ∞): f(x) = x² + 4. Show that inverse…
Let $R^+$ be the set of all positive real numbers and $f: R^+ \to [4, \infty): f(x) = x^2 + 4$. Show that inverse of $f$ exists and find $f^{-1}$.
Given: Let $R^+$ be the set of all positive real numbers and $f: R^+ \to [4, \infty): f(x) = x^2 + 4$
To show: Inverse of $f$ exists and find $f^{-1}$
Answer
Answer
Show $f$ is one-one and onto, then find $f^{-1}$.
One-one: $f(x_1) = f(x_2) \Rightarrow x_1^2 + 4 = x_2^2 + 4 \Rightarrow x_1^2 = x_2^2 \Rightarrow x_1 = x_2$ (both positive).
Onto: let $y = x^2 + 4$, $y \ge 4$; then $x = \sqrt{y-4}$ is a real number, so every $y$ in the codomain has a pre-image.
So $f$ is one-one and onto (as the question intends), hence $f^{-1}$ exists.
Hence $f^{-1}: [4, \infty) \to R^+$, $f^{-1}(y) = \sqrt{y-4}$.
- Let $x_1, x_2 \in R^+$ with $f(x_1) = f(x_2)$, by definition of one-one.
- $x_1^2 + 4 = x_2^2 + 4 \Rightarrow x_1^2 = x_2^2$.
- $x_1 = x_2$ since both are positive, so $f$ is one-one.
- Let $y \in [4, \infty)$ and put $y = x^2 + 4$, by definition of onto.
- $x = \sqrt{y-4}$ (positive root), which is real as $y \ge 4$, so $f$ is onto.
- $f$ is one-one and onto, so $f^{-1}$ exists.
- Hence $f^{-1}(y) = \sqrt{y-4}$, $y \in [4, \infty)$.