Which functions are the same as their inverse functions? !

Case a)
f(x)=[x-1]/[x+5]
step 1
f(x)=y
y=[x-1]/[x+5]
step 2
exchange x for y and y for x
y=[x-1]/[x+5]> x=[y-1]/[y+5]> x*[y+5]=[y-1]> xy+5x=y-1
step 3
clear the variable y
xy+5x=y-1> y-xy=5x+1> y*[1-x]=[5x+1]> y=[5x+1]/[1-x]
step 4
f(x)-1= [5x+1]/[1-x]
the function and the inverse function are not the same

case b)
g(x)=[x-2]/[x-1]
step 1
g(x)=y
y=[x-2]/[x-1]
step 2
exchange x for y and y for x
y=[x-2]/[x-1]> x=[y-2]/[y-1]> x*[y-1]=[y-2]> xy-x=y-2
step 3 
clear the variable y
xy-x=y-2> xy-y=-2+x> y*[x-1]=[x-2]> y=[x-2]/[x-1]
step 4
g(x)-1= [x-2]/[x-1]
the function and the inverse function are  the same

case c)
h(x)=[x+3]/[x-2]
step 1
h(x)=y
y=[x+3]/[x-2]
step 2
exchange x for y and y for x
y=[x+3]/[x-2]> x=[y+3]/[y-2]> x*[y-2]=[y+3]> xy-2x=y+3
step 3 
clear the variable y
xy-2x=y+3> xy-y=3+2x> y*[x-1]=[2x+3]> y=[2x+3]/[x-1]
step 4
h(x)-1= [2x+3]/[x-1]
the function and the inverse function are not the same

case d)
k(x)=[x+1]/[x-1]
step 1
k(x)=y
y=[x+1]/[x-1]
step 2
exchange x for y and y for x
y=[x+1]/[x-1]> x=[y+1]/[y-1]---> x*[y-1]=[y+1]> xy-x=y+1
step 3 
clear the variable y
xy-x=y+1> xy-y=x+1> y*[x-1]=[x+1]> y=[x+1]/[x-1]
step 4
k(x)-1= [x+1]/[x-1]
the function and the inverse function are  the same

y=-x+1

Step-by-step explanation: