# Find the domain of each of the following real function.

Question:

Find the domain of each of the following real function.

(i) $f(x)=\frac{3 x+5}{x^{2}-9}$

(ii) $f(x)=\frac{2 x-3}{x^{2}+x-2}$

(iii) $f(x)=\frac{x^{2}-2 x+1}{x^{2}-8 x+12}$

(iv)  $f(x)=\frac{x^{3}-8}{x^{2}-1}$

Solution:

(i) Given: $f(x)=\frac{3 x+5}{x^{2}-9}$

Need to find: Where the functions are defined.

To find the domain of the function f(x) we need to equate the denominator to 0.

Therefore,

$x^{2}-9=0$

$\Rightarrow x^{2}=9$

$\Rightarrow^{x=\pm 3}$

It means that the denominator is zero when $x=3$ and $x=-3$

So, the domain of the function is the set of all the real numbers except $+3$ and $-3$.

The domain of the function, $\mathrm{D} \mathrm{f}(\mathrm{x})=(-\infty,-3) \cup(-3,3) \cup(3, \infty)$.

(ii) Given: $f(x)=\frac{2 x-3}{x^{2}+x-2}$

Need to find: Where the functions are defined.

To find the domain of the function f(x) we need to equate the denominator to 0.

Therefore

$x^{2}+x-2=0$

$\Rightarrow^{x^{2}+2 x-x-2}=0$

$\Rightarrow x(x+2)-1(x+2)=0$

$\Rightarrow(x+2)(x-1)=0$

$\Rightarrow^{x=-2} \& x=1$

It means that the denominator is zero when $x=1$ and $x=-2$

So, the domain of the function is the set of all the real numbers except 1 and $-2$.

The domain of the function, $D_{f(x)}=(-\infty,-2) \cup(-2,1) \cup(1, \infty)$.

(iii) Given: $f(x)=\frac{x^{2}-2 x+1}{x^{2}-8 x+12}$

Need to find: Where the functions are defined.

To find the domain of the function f(x) we need to equate the denominator to 0.

Therefore,

$x^{2}-8 x+12=0$

$\Rightarrow x^{2}-2 x-6 x+12=0$

$\Rightarrow x(x-2)-6(x-2)=0$

$\Rightarrow(x-2)(x-6)=0$

$\Rightarrow^{x=2} \&^{x=6}$

It means that the denominator is zero when x = 2 and x = 6

So, the domain of the function is the set of all the real numbers except 2 and 6 .

The domain of the function, $\mathrm{D}^{\mathrm{t}(x)}=(-\infty, 2) \cup(2,6) \cup(6, \infty)$.

(iv) Given: $f(x)=\frac{x^{3}-8}{x^{2}-1}$

Need to find: Where the functions are defined.

To find the domain of the function f(x) we need to equate the denominator to 0.

Therefore

$x^{2}-1=0$

$\Rightarrow x^{2}=1$

$\Rightarrow x=\pm 1$

It means that the denominator is zero when $x=-1$ and $x=1$

So, the domain of the function is the set of all the real numbers except $-1$ and $+1$.

The domain of the function, $D_{f(x)}=(-\infty,-1) \cup(-1,1) \cup(1, \infty)$.