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xz_007 [3.2K]
9 months ago
5

Given r(x) = startfraction 11 over (x minus 4) squared endfraction , which represents a domain restriction on r(x) and the corre

sponding inverse function?.
Mathematics
1 answer:
Dmitry_Shevchenko [17]9 months ago
3 0

The domain and inverse of the function

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

Domain = \mathbb{R} - \{4\},

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

What is a function?

A function from A to B is a rule that assigns to each element of A a unique element of B. A is called the domain of the function and B is called the codomain of the function.

There are different operations on functions like addition, subtraction, multiplication, division and composition of functions.

The given function is

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

r(x) is not defined if x - 4 = 0

r(x) is not defined for x = 4

Domain = \mathbb{R} - \{4\},

Where \mathbb{R} is the set of all real number

Let r(x) = y

\frac{11}{(x-4)^2} = y\\(x - 4)^2 = \frac{11}{y}\\x - 4 = \pm \sqrt{\frac{11}{y}}\\x = \pm\sqrt{\frac{11}{y}}  +4

<em />r^{-1}(x) = \pm \sqrt{\frac{11}{x}} + 4

To learn more about function, refer to the link:

brainly.com/question/22340031

#SPJ4

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