$$f(x)\quad =1+\sqrt { 1+x } $$
$$y\quad =1+\sqrt { 1+x } $$
$$y^{ 2 }\quad =1+1+x$$
$$y^{ 2 }-2\quad =x$$
How is it $x=y^{ 2 }-2y$ ?
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asked Sep 9, 2014 at 17:09
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1$\begingroup$ Let $\sqrt {1+x}=z$ then $y=1+z$ does not imply $y^2=1+z^2$ but $y^2=(1+z)^2=1+2z+z^2$. Another approach would use $(y-1)^2=z^2$. $\endgroup$– Mark BennetSep 9, 2014 at 17:13
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$\begingroup$ $y^2=1+2\sqrt{1+x}+1+x$. $\endgroup$– Marc BogaertsSep 9, 2014 at 17:14
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1$\begingroup$ I think you want $x=y^2-2y$ at the end $\endgroup$– Mark BennetSep 9, 2014 at 17:14
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2 Answers
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As you have said $$y = 1 + \sqrt {1 + x} $$ so that $$x = {\left( {y - 1} \right)^2} - 1$$by interchanging $x$ and $y$, we get $$y = {\left( {x - 1} \right)^2} - 1 = {x^2} - 2x$$ which is the inverse of the initial function. We could check the answer by noting that if $g$ is the inverse of $f$, then $gof$ should be the unity operator. Now $f(x) = 1 + \sqrt {1 + x} $ and $g(x) = {x^2} - 2x$ so that $$\displaylines{ g(f(x)) = f{(x)^2} - 2f(x) \cr = {\left( {1 + \sqrt {1 + x} } \right)^2} - 2\left( {1 + \sqrt {1 + x} } \right) \cr = 1 + 2\sqrt {1 + x} + 1 + x - 2 - 2\sqrt {1 + x} \cr = x \cr} $$which confirms our answer.
answered Sep 9, 2014 at 17:19
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$$y\quad =1+\sqrt { 1+x }\iff y-1 = \sqrt{1+x}$$ $$ \iff (y-1)^2 = 1+x $$ $$\iff y^2 -2y + 1 = 1 + x $$ $$\iff y^2 - 2y = x$$
Your error is introduced when you attempted to square the right-hand side: $$(1 +\sqrt{x+1})^2 = 1 + 2\sqrt{x+1} + x+1 \neq 2 + x$$
This is just a straight-forward use of the fact that $$(a + b)^2 = a^2 + 2ab + b^2 \neq a^2 + b^2$$