Question

Given r(x) = StartFraction 11 Over (x minus 4) squared EndFraction , which represents a domain restriction on r(x) and the corresponding inverse function? X > 4; r–1(x) = 4 minus StartRoot StartFraction 11 Over x EndFraction EndRoot x > 4; r–1(x) = 4 + StartRoot StartFraction 11 Over x EndFraction EndRoot x > –4; r–1(x) = 4 minus StartRoot StartFraction 11 Over x EndFraction EndRoot x > –4; r–1(x) = 4 + StartRoot StartFraction 11 Over x EndFraction EndRoot

Answer 1

Answer:

x > 4; r–1(x) = 4 + StartRoot StartFraction 11 Over x EndFraction EndRoot

Step-by-step explanation:

We are given that

$r(x)=\frac{11}{(x-4)^2}$

We have to find the domain of restriction on r(x) and corresponding inverse function.

Let

$y=r(x)=\frac{11}{(x-4)^2}$

$(x-4)^2=\frac{11}{y}$

$x-4=\sqrt{\frac{11}{y}}$

$x=\sqrt{\frac{11}{y}}+4$

$r^{-1}(y)=\sqrt{\frac{11}{y}}+4$

Now, replace x by y and y by x then, we get

$r^{-1}(x)=\sqrt{\frac{11}{x}}+4$

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

$r^{-1}(x)=\sqrt{\frac{11}{x}}+4$

Option:

x > 4; r–1(x) = 4 + StartRoot StartFraction 11 Over x EndFraction EndRoot

Answer 2

Answer:

the answer is b

Step-by-step explanation:

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