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The function $f$ satisfies \[f(\sqrt{x + 1}) = \frac{1}{x}\]for all $x \ge -1,$ $x\neq 0.$ Find $f(2)$.

 Jun 2, 2020
 #1
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f(2) = 1/4.

 Jun 2, 2020
 #2
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ples halp

\(\text{The function $f$ satisfies $f(\sqrt{x + 1}) = \dfrac{1}{x}$ for all $x \ge -1,$ $x\neq 0.$ Find $f(2)$}. \)

 

\(\begin{array}{|rcll|} \hline \sqrt{x + 1} &=& 2 \\ x+1 &=& 4 \\ x &=& 4-1 \\ \mathbf{x} &=& 3 \\\\ f(\sqrt{x + 1}) &=& \dfrac{1}{x} \quad | \quad x = 3 \\\\ f(\sqrt{3 + 1}) &=& \dfrac{1}{3} \\\\ f(\sqrt{4}) &=& \dfrac{1}{3} \\\\ \mathbf{f(2)} &=& \mathbf{\dfrac{1}{3}} \\ \hline \end{array}\)

 

laugh

 Jun 2, 2020
edited by heureka  Jun 2, 2020
 #3
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thank you so much you are the best


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