Question

Find the domain and range of $f\left(x\right)=\frac{1}{\sqrt{2x-3}}$.

• A. Domain of $f\left(x\right)=(\frac{1}{2},\infty )$Range of $f\left(x\right)=[\frac{1}{2},\infty )$
• B. Domain of $f\left(x\right)=(\frac{3}{2},\infty )$Range of $f\left(x\right)=[\frac{3}{2},\infty )$
• C. Domain of $f\left(x\right)=(\frac{3}{2},\infty )$Range of $f\left(x\right)=[\frac{1}{2},\infty )$
• D. Domain of $f\left(x\right)=(\frac{1}{2},\infty )$Range of $f\left(x\right)=[\frac{3}{2},\infty )$

Answer 1

The correct option is B

Domain of $f\left(x\right)=(\frac{3}{2},\infty )$

Range of $f\left(x\right)=[\frac{3}{2},\infty )$

Given function,

$f\left(x\right)=\frac{1}{\sqrt{2x-3}}$

As we know $2x-3>0⟹x>\frac{3}{2}$

Domain of $f\left(x\right)=(\frac{3}{2},\infty )$

Let $f\left(x\right)=y$

$⟹y=\sqrt{2x-3}⟹2x-3=y^2⟹2x=y^2+3$

$⟹x=\frac{y^2+3}{2}$

As we know,

$y^2\ge 0⟹y^2+3\ge 3⟹\frac{y^2+3}{2}\ge \frac{3}{2}$

Range of $f\left(x\right)=[\frac{3}{2},\infty )$