Respuesta :

close enough, let's do the switcharoo of the variables and solve for "y".

[tex]\bf y=-2x+1\qquad \boxed{x}=-2\boxed{y}+1\implies x-1=-2y \\\\\\ \cfrac{x-1}{-2}=y\implies \cfrac{1-x}{2}=\stackrel{f^{-1}}{y}[/tex]
highwireact
So, again, we have our rules for mapping [tex]y[/tex] to [tex]x[/tex]:

[tex]y=-2x+1[/tex]

To find [tex]f^{-1}[/tex], we solve for [tex]x[/tex].

[tex]y-1=(-2x+1)-1[/tex] (subtract 1 from both sides)
[tex](y-1)/(-2)=(-2x)/(-2)[/tex] (divide both sides by -2)

That gives us

[tex]x= \frac{y-1}{-2} [/tex]

Now that we have a way of mapping [tex]y[/tex] back to [tex]x[/tex], all we do is swap the domain and range, and we have

[tex]f^{-1}(x)= \frac{x-1}{-2} [/tex]

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