Answer to Question #192330 in Algebra for Belden Ago

Question #192330

How to work out the domain and range for these equations


1. f(x)=√x-1


2. f(x)=6x-1/1-2x


3. f^-1(x)=(X+1)/(2x+6)


1
Expert's answer
2021-05-14T05:19:03-0400

1). "f(x) = \\sqrt{x-1}"


square root of any variable always takes positive values for real values so ,

"\\implies x-1\\ge0"

"\\implies x\\ge1"

so the domain is : "[1,+\\infty)"

by putting the values of domain in the function "f(x) = \\sqrt{x-1}" we get the range.

so the range is : "[0,+\\infty)"


2). "f(x)=\\frac{6x-1}{1-2x}"


To find the excluded value in the domain of the function, equate the denominator to zero and solve for x.

"\\implies1-2x=0"

"\\implies x=\\frac{1}{2}"


So, the domain of the function is set of real numbers except "\\frac{1}{2}" .

so the domain is : "(-\\infty,\\frac{1}{2})\\cup(\\frac{1}{2},+\\infty)"


The range of the function is same as the domain of the inverse function. So to find the range define the inverse of the function.


let, "f(x) = y = \\frac{6x-1}{1-2x}"

interchange x and y.


"x=\\frac{6y-1}{1-2y}"

solving for y we get,

x-2xy=6y-1


"y=\\frac{x+1}{6+2x}"


so, the inverse function is

"f^{-1}(x)=\\frac{x+1}{6+2x}"


the excluded value in he domain of the inverse function can be determined by equating the denominator to zero and solving for x.

6+2x = 0

x= -3

So, the domain of the inverse function is the set of real numbers except -3.

that is, the range of given function is the set of real numbers except -3.

So, the range is : "(-\\infty,-3)\\cup(-3,+\\infty)"


3). "f^{-1}(x)=\\frac{x+1}{2x+6}"


from above problem's solution we can see that the domain of the given function is :

6+2x = 0

x= -3

So, the domain of the inverse function is the set of real numbers except -3.

domain : "(-\\infty,-3)\\cup(-3,+\\infty)"


The range of the function is same as the domain of the inverse function. So to find the range define the inverse of the function.

"f(x)=\\frac{6x-1}{1-2x}"

domain of this inverse function is the range of the given function "f^{-1}(x)=\\frac{x+1}{2x+6}"

That is, 1-2x=0

x= 1/2

So, the domain of the inverse function is set of real numbers except "\\frac{1}{2}" .

that is, the range of given function is the set of real numbers except "\\frac{1}{2}" .

So, the range is : "(-\\infty,\\frac{1}{2})\\cup(\\frac{1}{2},+\\infty)" .



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