Question

Write the domain and range of fution f(x)=(x_2)/(x_3)​

Answer 1

Given,

$\longrightarrow f(x)=\dfrac{x-2}{x-3}$

Let us find its domain.

We know the denominator of a fraction cannot be zero.

$\longrightarrow x-3\neq0$

$\longrightarrow x\neq3$

Hence the domain is,

$\longrightarrow\underline{\underline{x\in\mathbb{R}-\{3\}}}$

Now let us find its range.

So,

$\longrightarrow f(x)=\dfrac{x-2}{x-3}$

$\longrightarrow (x-3)\,f(x)=x-2$

$\longrightarrow x\,f(x)-3\,f(x)=x-2$

$\longrightarrow x\,f(x)-x=3\,f(x)-2$

$\longrightarrow x(f(x)-1)=3\,f(x)-2$

$\longrightarrow x=\dfrac{3\,f(x)-2}{f(x)-1}$

As we said, the denominator of a fraction cannot be zero.

$\longrightarrow f(x)-1\neq0$

$\longrightarrow f(x)\neq1$

Hence the range is,

$\longrightarrow\underline{\underline{f(x)\in\mathbb{R}-\{1\}}}$

Answer 2

Answer:

Given,

\longrightarrow f(x)=\dfrac{x-2}{x-3}⟶f(x)=

x−3

x−2

Let us find its domain.

We know the denominator of a fraction cannot be zero.

\longrightarrow x-3\neq0⟶x−3



=0

\longrightarrow x\neq3⟶x



=3

Hence the domain is,

\longrightarrow\underline{\underline{x\in\mathbb{R}-\{3\}}}⟶

x∈R−{3}

Now let us find its range.

So,

\longrightarrow f(x)=\dfrac{x-2}{x-3}⟶f(x)=

x−3

x−2

\longrightarrow (x-3)\,f(x)=x-2⟶(x−3)f(x)=x−2

\longrightarrow x\,f(x)-3\,f(x)=x-2⟶xf(x)−3f(x)=x−2

\longrightarrow x\,f(x)-x=3\,f(x)-2⟶xf(x)−x=3f(x)−2

\longrightarrow x(f(x)-1)=3\,f(x)-2⟶x(f(x)−1)=3f(x)−2

\longrightarrow x=\dfrac{3\,f(x)-2}{f(x)-1}⟶x=

f(x)−1

3f(x)−2

As we said, the denominator of a fraction cannot be zero.

\longrightarrow f(x)-1\neq0⟶f(x)−1



=0

\longrightarrow f(x)\neq1⟶f(x)



=1

Hence the range is,

\longrightarrow\underline{\underline{f(x)\in\mathbb{R}-\{1\}}}⟶

f(x)∈R−{1}