Question

If $f:[\frac{1}{2},\infty )\to [\frac{3}{4},\infty )$ given by $f\left(x\right)=x^2-x+1,$ then $f^{-1}\left(x\right)=$

• A. $\frac{1}{2}+\sqrt{x-\frac{1}{4}}$
• B. $\frac{1}{4}+\sqrt{x-\frac{1}{2}}$
• C. $\frac{1}{2}+\sqrt{x-\frac{3}{4}}$
• D. $\frac{1}{2}-\sqrt{x-\frac{3}{4}}$

Answer 1

The correct option is C $\frac{1}{2}+\sqrt{x-\frac{3}{4}}$

$y=x^2-x+1\Rightarrow x^2-x+1-y=0\Rightarrow x=\frac{1\pm \sqrt{1-4(1-y)}}{2}$
Now replace $x$ with $y$, we get
$y=\frac{1\pm \sqrt{1-4(1-x)}}{2}$
We can take only the $+$ sign since the domain of $f\left(x\right)$ given as $[\frac{1}{2},\infty )$, so this is the range of $f^{-1}\left(x\right)$

$\therefore f^{-1}\left(x\right)=\frac{1}{2}+\sqrt{x-\frac{3}{4}}$