Bunuel wrote:
If the function \(f(x) =x^2+1\) and \(f(\sqrt{y+1})=1\), what is the value \(y\)?
(A) -4
(B) -1
(C) 1
(D) 3
(E) 5
APPROACH #1: Number sense
Given: \(f(x) =x^2+1\)
In other words, \(f(something) = something^2+1\)
So, if \(f(something) = 1\), we can be certain that \(something^2 = 0\)
Similarly, since we're told \(f(\sqrt{y+1})=1\), we can be certain that \((\sqrt{y+1})^2 = 0\), which means \(y = -1\)
Answer: B
APPROACH #2: Algebra
If \(f(x) =x^2+1\), then \(f(\sqrt{y+1})= (\sqrt{y+1})^2 + 1\)
Since we're told \(f(\sqrt{y+1})=1\), we can write: \(f(\sqrt{y+1})= (\sqrt{y+1})^2 + 1 = 1\)
Take: \((\sqrt{y+1})^2 + 1 = 1\)
Subtract \(1\) from both sides: \((\sqrt{y+1})^2 = 0\)
Simplify: \(y+1 = 0\)
Solve: \(y = -1\)
Answer: B
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