Additional Mathematics – Functions | e-Consult

Answer 1

• Domain of \(f\): solve \(2x-5\ge 0\Rightarrow x\ge \dfrac{5}{2}\). Hence \(\displaystyle \text{Dom}(f)=\left[\frac{5}{2},\infty\right)\).
• Range of \(f\): the square‑root yields non‑negative values, and as \(x\to\infty\), \(\sqrt{2x-5}\to\infty\). Thus \(\displaystyle \text{Ran}(f)= [0,\infty)\).
• Inverse: \(y=\sqrt{2x-5}\Rightarrow y^{2}=2x-5\Rightarrow x=\dfrac{y^{2}+5}{2}\). Hence \(f^{-1}(y)=\dfrac{y^{2}+5}{2}\) with \(y\ge0\). Therefore the domain of \(f^{-1}\) is the range of \(f\): \(\displaystyle \text{Dom}(f^{-1})=[0,\infty)\).