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

Oct 25, 2022

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Hi Guest!

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

To find $$f(2)$$ we need to make whatever inside the bracket of "f" to be 2.

I.e.

$$\sqrt{x+1}=2\\ \iff x+1=4 \implies x=3$$

So, if we substitute x=3 in the function we get the answer.

That is, $$f(\sqrt{3+1})=f(\sqrt{4})=f(2)=\dfrac{1}{3+1}=\dfrac{1}{4}$$which is the desired answer.

I hope this helps, and don't hesitate to ask for further clarificiation or any other question :) .

Oct 26, 2022