Answer :
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) :( -∞ , +∞)
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) :( -∞ , +∞)
To find the inverse function you need to change [tex]f(x)[/tex] (call it [tex]y[/tex]) and [tex]x[/tex], then solve for [tex]y[/tex]:
[tex]y = \frac{1}{2}x+3 \\ x = \frac{1}{2}y + 3 \\ x - 3 = \frac{1}{2}y \\ 2x-6 = y[/tex]
So now you have [tex]f^{-1}(x) = 2x-6[/tex].
Composition to prove inverse relation: [tex]f \circ f^{-1} (x) = x[/tex]:
[tex]f(f^{-1}(x)) = \frac{1}{2}(2x-6)+3 = x - 3 + 3 = x \square[/tex]
Domain and Range of both functions is Real numbers since they are both linear equations.
[tex]y = \frac{1}{2}x+3 \\ x = \frac{1}{2}y + 3 \\ x - 3 = \frac{1}{2}y \\ 2x-6 = y[/tex]
So now you have [tex]f^{-1}(x) = 2x-6[/tex].
Composition to prove inverse relation: [tex]f \circ f^{-1} (x) = x[/tex]:
[tex]f(f^{-1}(x)) = \frac{1}{2}(2x-6)+3 = x - 3 + 3 = x \square[/tex]
Domain and Range of both functions is Real numbers since they are both linear equations.
