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Nonlinear Functions

SAT Math· difficulty 3/5

What is the inverse of f(x)=x+2f(x) = \sqrt{x} + 2 (for x≥0x \geq 0)?

  • A

    f−1(x)=(x−2)2f^{-1}(x) = (x - 2)^2, x≥2x \geq 2

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  • B

    f−1(x)=x−2f^{-1}(x) = \sqrt{x - 2}

  • C

    f−1(x)=x2−2f^{-1}(x) = x^2 - 2

  • D

    f−1(x)=(x+2)2f^{-1}(x) = (x + 2)^2

Explanation

y=x+2⇒y−2=x⇒x=(y−2)2y = \sqrt{x} + 2 \Rightarrow y - 2 = \sqrt{x} \Rightarrow x = (y - 2)^2. So f−1(x)=(x−2)2f^{-1}(x) = (x - 2)^2 for x≥2x \geq 2.

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