Answer:
1. [tex](f^{-1}og^{-1}) (x) = \sqrt{x^{2}+3 }[/tex]
2. [tex](f o g)^{-1} (x) = x + 3[/tex]
Step-by-step explanation:
f(x) = [tex]x^{2}[/tex] + 1
putting f (x) = y
y = [tex]x^{2}[/tex] + 1
[tex]x^{2}[/tex] = ( y-1)
x = [tex](y - 1)^{\frac{1}{2} }[/tex]
Therefore = [tex]f^{-1}(x)[/tex] = [tex](x-1)^{\frac{1}{2} }[/tex]
g (x) = [tex]\sqrt{x-4}[/tex]
putting g(x) = y
y = [tex]\sqrt{x-4}[/tex]
[tex]y^{2}[/tex] = (x-4)
x = [tex]y^{2} + 4[/tex]
[tex]g^{-1} (x) = x^{2} + 4[/tex]
( [tex]f^{-1}[/tex] o [tex]g^{-1}) x[/tex]
[tex](g^{-1}x -1)^{\frac{1}{2} } = \sqrt{x^{2} +3}[/tex]
2. (f o g)(x) = (√x-4)²+1) = x-4+1
(f o g)(x) = x-3
Let y = x - 3
x = y + 3
Or [tex](f o g)^{-1}(x) = x + 3[/tex]
