Answer:
[tex]f^{-1}(x)=-2x+6[/tex].
Step-by-step explanation:
[tex]y=f(x)[/tex]
[tex]y=3-\frac{1}{2}x[/tex]
The biggest thing about finding the inverse is swapping x and y. The inverse comes from switching all the points on the graph of the original. So a point (x,y) on the original becomes (y,x) on the original's inverse.
Sway x and y in:
[tex]y=3-\frac{1}{2}x[/tex]
[tex]x=3-\frac{1}{2}y[/tex]
Now we want to remake y the subject (that is solve for y):
Subtract 3 on both sides:
[tex]x-3=-\frac{1}{2}y[/tex]
Multiply both sides by -2:
[tex]-2(x-3)=y[/tex]
We could leave as this or we could distribute:
[tex]-2x+6=y[/tex]
The inverse equations is [tex]y=-2x+6[/tex].
Now some people rename this [tex]f^{-1}[/tex] or just call it another name like [tex]g[/tex].
[tex]f^{-1}(x)=-2x+6[/tex].
Let's verify this is the inverse.
If they are inverses then you will have that:
[tex]f(f^{-1}(x))=x \text{ and } f^{-1}(f(x))=x[/tex]
Let's try the first:
[tex]f(f^{-1}(x))[/tex]
[tex]f(-2x+6)[/tex] (Replace inverse f with -2x+6 since we had [tex]f^{-1})(x)=-2x+6[/tex]
[tex]3-\frac{1}{2}(-2x+6)[/tex] (Replace old output, x, in f with new input, -2x+6)
[tex]3+x-3[/tex] (I distributed)
[tex]x[/tex]
Bingo!
Let's try the other way.
[tex]f^{-1}(f(x))[/tex]
[tex]f^{-1}(3-\frac{1}{2}x)[/tex] (Replace f(x) with 3-(1/2)x since [tex]f(x)=3-\frac{1}{2}x[/tex])
[tex]-2(3-\frac{1}{2}x)+6[/tex] (Replace old input, x, in -2x+6 with 3-(1/2)x since [tex]f(x)=3-\frac{1}{2}x[/tex])
[tex]-6+x+6[/tex] (I distributed)
[tex]x[/tex]
So both ways we got x.
We have confirmed what we found is the inverse of the original function.
