Question

The function f is such that f(x) = x^2 - 8x +5 Where x<=4 express the inverse function f-1 in the form f-1(x)

Answer 1

`f^(-1)`(x) =
`√(x+11)` + 4, x=-11

Explanation:

y = x^2 - 8x + 5, x<=4

x = y^2 - 8y + 5, y<=4 ==> switch the x and y variables to find the inverse

x = y^2 - 8y + 5 ==> solve for y

x + 16 = y^2 - 8y + 16 + 5 ==> add 16 to get a perfect polynomial square

x + 16 = (y - 8/2)^2 + 5 ==> simplify

x + 16 = (y - 4)^2 + 5

x + 16 - 5 = (y - 4)^2 + 5 - 5 ==> isolate y by subtracting 5 on both sides

x + 11 = (y - 4)^2 ==> simplify

`√(x+11)` =
`√((y - 4)^2)` ==> take the square root of both sides to remove the

square

`√(x+11)` = y - 4

y =
`√(x+11)` + 4 ==> add 4 on both sides to isolate y

`f^(-1)`(x) =
`√(x+11)` + 4 ==> substitute
`f^(-1)`(x) for y

`f^(-1)`(x) =
`√(x+11)` + 4,
`f^(-1)`(x)<=4 ==> add in the domain restriction

4 =
`√(x+11)` + 4 ==> plugin the domain restriction for
`f^(-1)`(x)

0 =
`√(x+11)` ==> subtract 4 on both sides

x + 11 = 0 ==> the square root of 0 is 0

x = -11 ==> subtract 11 on both sides

Hence, answer is:
`f^(-1)`(x) =
`√(x+11)` + 4, x=-11