The function f is defined for x \geq 1 by f(x) = (x - 1)^2. What is f^{-1}(x)?

f^{-1}(x) = \sqrt{x} + 1 y = (x - 1)^2 \Rightarrow \sqrt{y} = x - 1 (taking positive root since x \geq 1) \Rightarrow x = \sqrt{y} + 1. So f^{-1}(x) = \sqrt{x} + 1.

Nonlinear Functions

SAT Math· difficulty 5/5

f

A
$f^{-1}(x) = \sqrt{x} + 1$
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B
$f^{-1}(x) = (x + 1)^2$
C
$f^{-1}(x) = \sqrt{x} - 1$
D
$f^{-1}(x) = \sqrt{x - 1}$

Explanation

y = (x - 1)^2 \Rightarrow \sqrt{y} = x - 1

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