Question

If f(x) = $\frac{x - 1}{x + 1}$, x ≠ ± 1, then verify fof⁻¹ (x) = x.

Answer 1

Explanation:

This is a question of  "Composition of Functions"   &

"Inverse Functions".

Composition of Functions:

Let f be a function from set X to set Y and g be a function from set Y to set Z. The composition of f & g is a function, denoted by gof.

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Answer 2

As we know that while finding inverse of the function ,it must be written in terms of y.

ie f(y) = x

here

$y = f(x) \\ \\ f(x) = \frac{x - 1}{x + 1} \\ \\ y = \frac{x - 1}{x + 1} \\ \\ yx + y = x - 1 \\ \\ yx - x = - 1 - y \\ \\ x(y - 1) = - 1 - y \\ \\ x(1 - y) = 1 + y \\ \\ x = \frac{1 + y}{1 - y} \\ \\ {f}^{ - 1} = \frac{1 + y}{1 - y} \\ \\ {f}^{ - 1} (x)= \frac{1 + x}{1 - x}$
now for

$fo {f}^{ - 1} (x)$
put the value of f inverse in the f(x)
$fo {f}^{ - 1} x = \frac{ \frac{1 + x}{1 - x} - 1 }{ \frac{1 + x}{1 - x} + 1} \\ \\ \\ = \frac{ \frac{1 +x - 1 +x}{1 - x} }{ \frac{1 +x + 1 - x}{1 + x}} \\ \\ \\ = \frac{2x}{2} \\ \\ = x$