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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:

Sagot :

subediprashant
 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) :( -∞ , +∞)
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.
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