Question

Let $f:[2,\infty )\to [0,\infty )$ defined by $f\left(x\right)=x^2-4x+4$, then $f^{-1}\left(x\right)$ is equal to

• A. $2+\sqrt{x}$
• B. $2-\sqrt{x}$
• C. $\sqrt{x}-2$
• D. $\sqrt{x}$

Answer 1

The correct option is A $2+\sqrt{x}$

Given : $f\left(x\right)=x^2-4x+4$
$f\left(x\right)=(x-2{)}^2$
Clearly it is one one and onto for $f:[2,\infty )\to [0,\infty )$, so
$y=(x-2{)}^2\Rightarrow x=2\pm \sqrt{y}\Rightarrow x=2+\sqrt{y}(\because x\in [2,\infty ))\therefore f^{-1}(x)=2+\sqrt{x}$