exchange x and y

x = (y + 2) / (5y – 1)

multiply 5y – 1 for both sides

x(5y – 1) = y + 2

distribute

5xy – x = y + 2

subtract y for both sides

5xy – x – y = 2

add x for both sides

5xy – y = x + 2

factor out y

y(5x – 1) = (x + 2)

divide 5x – 1

y = (x + 2) / (5x – 1)

in this case, the orgional function is inverse of itself.

f^-1(x) = (x + 2) / (5x – 1)

y = x^2 + 2x – 1

exchange x and y

x = y^2 + 2y – 1

add 1 for both sides

x + 1 = y^2 + 2y

add (b/2)^2 for both sides

x + 1 + (2/2)^2 = y^2 + 2y + (2/2)^2

x + 2 = y^2 + 2y + 1

factor

x + 2 = (y + 1)^2

take a square root

+/- sqrt(x + 2) = y + 1

subtract 1 for both sides

y = +/- sqrt(x + 2) – 1

since the domain is x > 0, take the positive one

f^-1(x) = sqrt(x + 2) – 1

2) put y in place of x and vice versa

3) solve for y

4) replace f(x) for y

a) x=(y+2)/(5y-1)

x(5y-1)=y+2

5xy-x=y+2

y-5xy= -(x+2)

y(1-5x)= -(x+2)

y= -(x+2)/(1-5x)

f^-1(x)=(x+2)/(5x-1)

b) x=y^2+2y-1

x+1=y^2 +2y

additional step: add (2/2)^2 for both sides

x+2= y^2 + 2y + 1

x+2=(y+1)^2

+/- sqrt (x+2)=y+1

since the domain is x>0 for the original function, the inverse function has a range y>0

so take the positive:

sqrt (x+2)=y+1

f^-1(x) = sqrt(x + 2) – 1