Question 115

If $$f(x) = \frac{x - 1}{x + 2}$$; x ≠ -2, then $$f^{-1}(4) =$$

Solution

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$$f^'\left(x\right)=\frac{\left(\left(x-2\right)1\ -\left(x-1\right)1\right)}{\left(x-2\right)^2}=$$

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