Answer: [tex]g^{-1} (8)=-3[/tex]
[tex]h^{-1}(x)=\dfrac{x-2}{3}[/tex]
[tex](h^{-1} \circ h)(-3)=-3[/tex].
Step-by-step explanation:
Given: The one-to-one functions g and h are defined as follows.
g=((-5, 2),( -3, 8), (-1, - 8), (8, 9)) [here each x= input values , y= output values in (x,y)]
h(x)=3x+2
To find: [tex]g^{-1} (8)[/tex]
As for 8 is a image of -3 ( from point ( -3, 8)).
So, [tex]g^{-1} (8)=-3[/tex]
To find : [tex]h^{-1} (x)[/tex]
Let [tex]y = h(x)=3x+2[/tex] (i)
Then,
[tex]y=3x+2\Rightarrow\ 3x= y-2\\\\\Rightarrow\ x=\dfrac{y-2}{3}[/tex]
Switch [tex]y[/tex] with [tex]x[/tex] and [tex]x[/tex] with [tex]h^{-1} (x)[/tex], we get
[tex]h^{-1}(x)=\dfrac{x-2}{3}[/tex] (ii)
To find : [tex](h^{-1} \circ h)(-3)[/tex]
Consider [tex](h^{-1} \circ h)(-3)=h^{-1} (h(-3))[/tex]
[tex]=h^{-1}(3(-3)+2)\ \ (using\ (i))\\\\=h^{-1}(-9+2)\\\\=h^{-1}(-7)\\\\= \dfrac{-7-2}{3}\ \ (using\ (ii))\\\\=\dfrac{-9}{3}=-3[/tex]
So, [tex](h^{-1} \circ h)(-3)=-3[/tex].
