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  4. Let f(x)=(x+1)2-1, x ≥-1 then the set x: f(x)=f-1(x) is equal to

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Q. Let $f(x)=(x+1)^2-1, x \geq-1$ then the set $\left\{x: f(x)=f^{-1}(x)\right\}$ is equal to

Relations and Functions - Part 2

Solution:

Let $ y=(x+1)^2-1, x \geq-1$
$\Rightarrow (x+1)^2=y+1 $
$\Rightarrow x+1=\sqrt{y+1} \text { as } x \geq-1 $
$\Rightarrow x=-1+\sqrt{y+1}, y \geq-1 $
$\text { Thus } f^{-1}(x)=-1+\sqrt{x+1} $
$\text { So } f(x)=f^{-1}(x)$
$\Rightarrow (x+1)^2-1=-1+\sqrt{x+1}$
$\Rightarrow \sqrt{x+1}=0 \text { or }(x+1)^{3 / 2}=1 $
$\Rightarrow x=-1 \text { or } x=0 $