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). "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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