Find error in this algebraic transformation on a function

Question

Let $f(x)=x^2+x\sqrt{x^2-1}$. Apparently the statement

$\forall x$ for which $x,-x\in\mathbb D$, $f(x)=f(-x)$.

is false, but I can “prove” it by \begin{align}x^2+x\sqrt{x^2-1}=x\left(x+\sqrt{x^2-1}\right)=\frac x{x-\sqrt{x^2-1}}=\frac1{1-\sqrt{1-\frac1{x^2}}}.\end{align} I still don’t see any problems in it.

@Chazg I mean the set on which the function is defined. For the function $f(x)$ defined above, $\mathbb D=(-\infty,-1)\cup(1,\infty)$ — , Sep 30 '22 at 07:47

score 8 · Accepted Answer · answered Sep 30 '22 at 07:41

8

In last step, when you "pull out $x$" from the square root, you really should be pulling out $|x|$. Fixing this yields result $$ \frac 1{ 1-\operatorname{sgn}(x) \sqrt{1-\frac 1{x^2}} }. $$

answered Sep 30 '22 at 07:41

Esgeriath

2,181

Find error in this algebraic transformation on a function

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