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What's the inverse of ƒ(x) = x2 – 16?

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Wallind123idn

the inverse of ƒ(x) = x²– 16

[tex] {f}^{ - 1} (x) = \sqrt{x + 16} [/tex]

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CouragePabaiidn

Answer:

The inverse function should be f

[tex] {f}^{ - 1}( x) = \sqrt{x + 16} [/tex]

this is the correct answer

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