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Question
If f(x) = [4 – (x – 7)3]1/5 is a real invertible function, then find f–1(x).
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Solution
Given f(x) = [4 – (x – 7)3]1/5 is a real invertible function.
Let f(x) = y
`\implies` y = [4 – (x – 7)3]1/5
`\implies` y5 = 4 – (x – 7)3
`\implies` (x – 7)3 = 4 – y5
`\implies` x – 7 = [4 – y5]1/3
`\implies` x = 7 + (4 – y5)1/3
`\implies` f1(y) = 7 + (4 – y5)1/3 ` {{:(∵ f(x) "is invertible"), (∴ f(x) = "y"),(\implies x = f^-1("y")):}`
Hence f1(x) = 7 + (4 – x5)1/3
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