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Given function $$ f(x) = \frac{-2 + 2 \sqrt{x+1}}{x} $$ and $$ g(x) = \frac{2}{1+\sqrt{x+1}} $$ Prove that $f(x) = g(x)$ where $x\neq 0$.

This question is from a 2018 examen in the Netherlands. I tried rewriting f(x) as follows: $$ f(x) = \frac{-2 + 2 \sqrt{x+1} * \sqrt{x+1}}{x*\sqrt{x+1}} $$ $$ f(x) = \frac{-2 + 2(x+1)}{x*\sqrt{x+1}} $$ $$ f(x) = \frac{-2 + 2x + 2}{x*\sqrt{x+1}} $$ $$ f(x) = \frac{2x}{x*\sqrt{x+1}} $$ $$ f(x) = \frac{2}{\sqrt{x+1}} $$

However, I can't find a way to get to the $h(x)$ function where the denominator has a 1 + still.

EDIT: Just as I post this I notice that I made a mistake in the first step. By not multiplying against the whole numerator. I'll leave the question open anyways.

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    $\begingroup$ Start from $g(x)$, and multiply by $\frac{1-\sqrt{x+1}}{1-\sqrt{x+1}}$ to obtain $f(x)$. $\endgroup$ Mar 24, 2021 at 20:28

3 Answers 3

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You can start with $f(x)$ by doing the following:

$$ \begin{alignat}{1} f(x) &= \frac{-2 + 2 \sqrt{x+1}}{x} \cdot \frac{(-2-2\sqrt{x+1})}{(-2-2\sqrt{x+1})} = \frac{4-4(x+1)}{-2x-2x\sqrt{x+1}}=\frac{-4x}{-2x\cdot(1+\sqrt{x+1})} \\ &= \left(\frac{-2x}{-2x}\right)\cdot \left( \frac{2}{1+\sqrt{x+1}} \right) = \frac{2}{1+\sqrt{x+1}}=g(x) \end{alignat} $$


But what's the idea behind it?

When solving this type of problem, look at the structure of the functions you are working with. For the functions in your question:

$$ \newcommand{\altfrac}{\genfrac{}{}{0pt}{}} \begin{alignat}{1} f(x) = \frac{-2 + 2 \sqrt{x+1}}{x} &\altfrac{\leftarrow \text{Square root here}~~~~~}{\leftarrow \text{No square root here}} \\[8pt] g(x) = \frac{2}{1+\sqrt{x+1}} &\altfrac{\leftarrow \text{No square root here}}{\leftarrow \text{Square root here}~~~~~} \end{alignat} $$

Going from $f(x)$ to $g(x)$, you need to make the square root appear in the denominator, so a good start would be to rationalize the numerator of $f(x)$.

Similary, from $g(x)$ to $f(x)$, you need to make the square root appear in the numerator, so a good start would be to rationalize the denominator of $g(x)$, as done in @vitamin d answer.

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    $\begingroup$ Thanks for writing that up, I see that's a lot more difficult that the other way around. $\endgroup$ Mar 24, 2021 at 20:45
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Third binomial formula: $(a+b)(a-b)=a^2-b^2$ $$\frac{2}{1+\sqrt{x+1}}=\frac{2(1-\sqrt{x+1})}{(1+\sqrt{x+1})(1-\sqrt{x+1})}=\frac{-2 + 2 \sqrt{x+1}}{1^2-(\sqrt{x+1})^2}=\frac{-2 + 2 \sqrt{x+1}}{x}$$

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  • $\begingroup$ Could you elaborate a bit on your thought process. Why for example you multiply with (1-sqrt(x+1)). Is this just by practicing a lot that you see this? $\endgroup$ Mar 24, 2021 at 20:32
  • $\begingroup$ That is the usual procedure to remove square roots from the denominator. I'd say this is a "well-known step". $\endgroup$
    – vitamin d
    Mar 24, 2021 at 20:33
  • $\begingroup$ Oke thanks, what about starting with g(x) over starting with f(x), do you intuitively choose? $\endgroup$ Mar 24, 2021 at 20:34
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    $\begingroup$ @GertjanBrouwer The well-known step being referred to is called 'Rationalizing the denominator'. $\endgroup$ Mar 24, 2021 at 20:35
  • $\begingroup$ @GertjanBrouwer I'm sorry I'm not sure if I understood you correctly. What is your question? $\endgroup$
    – vitamin d
    Mar 24, 2021 at 20:37
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Hint: consider the following identity $$ (\sqrt{x+1} + 1)(\sqrt{x+1} - 1) = (\sqrt{x+1})^2 - 1^2 = x + 1 - 1 = x $$

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