Question

1) Find the inverse function of f(x)=1/2x+3

2)Use composition to verify that they are inverse relations?
3) f^ Domain : Range:
4) f^-1 Domain : Range:

Answer 1

To find the inverse function you need to change
`f(x)` (call it
`y`) and
`x`, then solve for
`y`:


`y = (1)/(2)x+3 \\ x = (1)/(2)y + 3 \\ x - 3 = (1)/(2)y \\ 2x-6 = y`

So now you have
`f^(-1)(x) = 2x-6`.

Composition to prove inverse relation:
`f \circ f^(-1) (x) = x`:


`f(f^(-1)(x)) = (1)/(2)(2x-6)+3 = x - 3 + 3 = x \square`

Domain and Range of both functions is Real numbers since they are both linear equations.

Answer 2

y = 1/2x + 3
change x and y

x = 1/2y + 3
now find the value of y and that is inverse function

x - 3 =1/2 y
y = 2x- 6

f-(x) = 2x - 6


for both domain : ( - ∞ , + ∞ )

range f (x) : ( - ∞ , + ∞ ) - { 0 }
range f-(x) :( -∞ , +∞)