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Question
Answer the following:
If f(x) = log(1 – x), 0 ≤ x < 1 show that `"f"(1/(1 + x))` = f(1 – x) – f(– x)
Solution
f(x) = log(1 – x)
Replacing x by `(1/(1 + x))`, we get
`"f"(1/(1 + x)) = log(1 - 1/(1 + x))`
= `log((1 + x - 1)/(1 + x))`
= `log(x/(1 + x))`
∴ `"f"(1/(1 + x))` = log x – log(1 + x)
∴ `"f"(1/(1 + x))` = log(1 – 1 + x) – log(1 + x)
∴ `"f"(1/(1 + x))` = log[1 – (1 – x)] – log[1 – (– x)]
∴ `"f"(1/(1 + x))` = f(1 – x) – f(– x)
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