Question

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)

Answer 1

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 : $(-\infty,\frac{1}{2})\cup(\frac{1}{2},+\infty)$ .