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