Respuesta :
Answer:
Given [tex]g(x)=\frac{-1}{2}x+5[/tex] we can conclude the following things:
A) [tex]g[/tex] is one-to-one.
B) [tex]g[/tex]'s inverse is given by [tex]g^{-1}(x)=-2(x-5)[/tex].
Step-by-step explanation:
A) If it is one-to-one, then f(a)=f(b) implies only a=b.
The function is [tex]g(x)=\frac{-1}{2}x+5[/tex].
Let's see what [tex]g(a)=g(b)[/tex] implies here.
[tex]g(a)=g(b)[/tex]
[tex]\frac{-1}{2}a+5=\frac{-1}{2}b+5[/tex]
Subtract 5 on both sides:
[tex]\frac{-1}{2}a=\frac{-1}{2}b[/tex]
Multiply both sides by -2:
[tex]a=b[/tex]
This is exactly what we need to show that [tex]g[/tex] is one-to-one.
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If you look at a visual of [tex]g[/tex] on a graph. You would see it passes the horizontal line test making it one-to-one. (It is is just a diagonal line after all.)
B)
[tex]g(x)=\frac{-1}{2}x+5[/tex]
[tex]y=\frac{-1}{2}x+5[/tex]
The inverse is the swapping of [tex]x[/tex] and [tex]y[/tex] and then we want to remake the new [tex]y[/tex] the subject of the equation.
That is we have to solve the following for [tex]y[/tex]:
[tex]x=\frac{-1}{2}y+5[/tex]
Subtract 5 on both sides:
[tex]x-5=\frac{-1}{2}y[/tex]
Multiply both sides by -2:
[tex]-2(x-5)=y[/tex]
[tex]y=-2(x-5)[/tex]
So [tex]g^{-1}(x)=-2(x-5)[/tex].
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Let's verify.
So we should get that [tex]g^{-1}(g(x))=x \text{ and } g^{-1}(g(x))=x[/tex].
Let's see what happens:
[tex]g^{-1}(g(x))[/tex]
[tex]g^{-1}(\frac{-1}{2}x+5)[/tex]
[tex]-2(\frac{-1}{2}x+5-5)[/tex]
[tex]-2(\frac{-1}{2}x+0)[/tex]
[tex]-2(\frac{-1}{2}x)[/tex]
[tex]1x[/tex]
[tex]x[/tex] -So that looks good so far.
[tex]g(g^{-1}(x))[/tex]
[tex]g(-2(x-5))[/tex]
[tex]\frac{-1}{2}(-2(x-5))+5[/tex]
[tex](x-5)+5[/tex]
[tex]x-5+5[/tex]
[tex]x+0[/tex]
[tex]x[/tex] -So that looks good.
Both ways check out so we indeed found the inverse of [tex]g[/tex].
