Question

find the domain and range of f(x)=1/2x-1​

Answer 1

Answer:

Step-by-step explanation:

So, the Above Equation can be written as,

“ y = x+1/2x+1 ” ………… (1)

So, We can put any Value for x except -1/2.

Therfore, Domain for the Above Equation will be,

= R - {-1/2}

Now, for Range we have to First Write the above Eqn. in terms of y

i.e. “x = 1 - y/2y - 1”

Here,We can put any value for y except 1/2.

Therefore, Range will be ,

= R - {1/2}

Here,

R = All Real Numbers

R - {-1/2} = All Real Numbers except -1/2.

Reason behind is, if we put -1/2 as value of “x” , it will produce ZERO in Denominator which makes the Equation as Not Defined.

R - {1/2} = All Real Numbers except 1/2.

Reason behind is, if we put 1/2 as value of “y” , it will produce ZERO in Denominator which makes the Equation as Not Defined.

Answer 2

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:f(x) = \dfrac{1}{2x - 1}$

We know

Domain of a function f(x) is defined as set of those values of x for which function f(x) is well defined.

So, Domain of f(x) is defined when

$\rm :\longmapsto\:2x - 1 \: \ne \: 0$

$\rm :\longmapsto\:2x \: \ne \: 1$

$\rm :\longmapsto\:x \: \ne \: \dfrac{1}{2}$

$\bf\implies \:Domain \: of \: f(x) : x \in \: R - \bigg \{\dfrac{1}{2} \bigg \}$

Now,

Range of a function f(x) is defined as set of those values of f(x) attain by f(x) on substituting x.

To find the range of f(x), Let assume that

$\rm :\longmapsto\:y = \dfrac{1}{2x - 1}$

$\rm :\longmapsto\:2x - 1 = \dfrac{1}{y}$

$\rm :\longmapsto\:2x = \dfrac{1}{y} + 1$

$\rm :\longmapsto\:2x = \dfrac{1 + y}{y}$

$\rm :\longmapsto\:x = \dfrac{1 + y}{2y}$

So, x is defined if

$\rm :\longmapsto\:2y \: \ne \: 0$

$\rm :\longmapsto\:y \: \ne \: 0$

$\bf\implies \:Range \: of \: f(x) : y \in \: R - \bigg \{0 \bigg \}$

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Basic Concept Used

Substitute y = f(x) and then solve the equation for x, and represents in the form x = g(y).

Find the domain of g(y), and this will be the range of f(x).