Answer:
x > 4; r–1(x) = 4 + StartRoot StartFraction 11 Over x EndFraction EndRoot
Step-by-step explanation:
We are given that
[tex]r(x)=\frac{11}{(x-4)^2}[/tex]
We have to find the domain of restriction on r(x) and corresponding inverse function.
Let
[tex]y=r(x)=\frac{11}{(x-4)^2}[/tex]
[tex](x-4)^2=\frac{11}{y}[/tex]
[tex]x-4=\sqrt{\frac{11}{y}}[/tex]
[tex]x=\sqrt{\frac{11}{y}}+4[/tex]
[tex]r^{-1}(y)=\sqrt{\frac{11}{y}}+4[/tex]
Now, replace x by y and y by x then, we get
[tex]r^{-1}(x)=\sqrt{\frac{11}{x}}+4[/tex]
The function is not defined at x=4
But,
The inverse function defines for all positive real values.
Therefore, domain of r(x)=x>4 and inverse function
[tex]r^{-1}(x)=\sqrt{\frac{11}{x}}+4[/tex]
Option:
x > 4; r–1(x) = 4 + StartRoot StartFraction 11 Over x EndFraction EndRoot
Answer:
the answer is b
Step-by-step explanation:
i took the test on edg
