Functions that are its own Inverse.

Question

Show that the function defined by is its own inverse for any values of and .

Answer 1

f(x) = (x+h)/(kx-1)

y= (x+h)/(kx-1) switch x and y

x =(y+h)/(ky-1) solve for y

xky -x = y+h

xky -y = x+h

y(kx-1) = x+h

y = (x+h)/(kx-1)

f^-1(x) = (x+h)/(kx-1) = f(x)

f(x) = (x+h)/(kx-1)

f(f(x)) = [(x+h)/(kx-1) + h]/[k(x+h)/(kx-1) -1]

((x+h+hkx-h)/(kx-1))/(kx+kh-kx+1)/(kx-1)

= (x+hkx)/(kh+1)= x(kh+1)/(kh+1)

=x

f(f(x))= x = f^-1(f(x)

f(x) = f^-1(x)