Question

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Answer 1

Answer:

$f[2-f^{-1}(x)]=x$

Step-by-step explanation:

Given function is,

f(x) = $\frac{x-1}{2}$

Equation form of the function will be,

y = $\frac{x-1}{2}$

Interchange x and y,

$x=\frac{y-1}{2}$

Solve for y,

y = 2x + 1

Therefore, inverse of the function 'f' will be,

$f^{-1}(x)=2x+1$

Now $[2-f^{-1}(x)]$ = 2 - [2x + 1]

                          = -2x + 1

Value of $f[2-f^{-1}(x)]$ = f(2x + 1)

From the given function,

f(2x + 1) = $\frac{(2x+1)-1}{2}$

            = x

Therefore, value of  $f[2-f^{-1}(x)]=x$