Question

Find domain and range of given equation $f\left(x\right)={\displaystyle \frac{3}{2-x^2}}$

Answer 1

Hint: We can find the domain by finding the value of x that makes the function undefined and subtracting it from the set of real numbers. For that we equate the denominator to 0 and solve for x. For finding the range, we can write the interval of $x^2$ . Then we can transform it to the function by addition, multiplication and taking reciprocals, to get the range of the function.

Complete step-by-step answer:
We know that domain of a function is the set of all the possible values that the variable in that function takes. We know that a real valued function is not defined when the denominator becomes zero and the term inside the root is negative.
Here, we have the function $f\left(x\right)={\displaystyle \frac{3}{2-x^2}}$ .
The function has a denominator. So, the range must not have the value which makes the denominator 0.
We can equate the denominator to 0 and solve it to get that value of x.
 $\Rightarrow 2-x^2=0$
On rearranging, we get,
 $\Rightarrow x^2=2$
On taking the square root, we get,
 $\Rightarrow x=\pm \sqrt{2}$
So, x cannot take the values $-\sqrt{2}$ and $\sqrt{2}$ .
Therefore, the domain of the function is set of all the real numbers except the numbers $-\sqrt{2}$ and $\sqrt{2}$ .
 $\Rightarrow D=R-\left\{-\sqrt{2},\sqrt{2}\right\}$
Range of a function is the set of all the values that the function gives for the values of variables in the domain.
 $f\left(x\right)={\displaystyle \frac{3}{2-x^2}}$
We know that, $\mathrm{\infty }>x^2⩾0$
On multiplying with -1, the equality reverses,
$$\Rightarrow$$ $-\mathrm{\infty }<-x^2⩽0$
On adding 2, we get,
$$\Rightarrow$$ $-\mathrm{\infty }<2-x^2⩽2$
According to the domain of the function, $x^2\ne 2$ . So $2-x^2$ cannot take value 0.
  $\Rightarrow -\mathrm{\infty }<2-x^2<0$ and $0<2-x^2<2$
On taking the reciprocal, we get,
 $\Rightarrow {\displaystyle \frac{1}{-\mathrm{\infty }}}>{\displaystyle \frac{1}{2-x^2}}>{\displaystyle \frac{1}{0}},{\displaystyle \frac{1}{0}}>{\displaystyle \frac{1}{2-x^2}}⩾{\displaystyle \frac{1}{2}}$
On multiply throughout with 3, we get,
 $\Rightarrow {\displaystyle \frac{3}{-\mathrm{\infty }}}>{\displaystyle \frac{3}{2-x^2}}>{\displaystyle \frac{3}{0}},{\displaystyle \frac{3}{0}}>{\displaystyle \frac{3}{2-x^2}}⩾{\displaystyle \frac{3}{2}}$
We know that ${\displaystyle \frac{1}{\mathrm{\infty }}}=0$ and ${\displaystyle \frac{1}{0}}=\mathrm{\infty }$
 $\Rightarrow 0>{\displaystyle \frac{3}{2-x^2}}>-\mathrm{\infty },\mathrm{\infty }>{\displaystyle \frac{3}{2-x^2}}⩾{\displaystyle \frac{3}{2}}$
So, y can take values in the interval $\left(-\mathrm{\infty },0\right)\cup \left[{\displaystyle \frac{3}{2}},\mathrm{\infty }\right)$
Therefore, the domain of the function is $R-\left\{-\sqrt{2},\sqrt{2}\right\}$ and range is $\left(-\mathrm{\infty },0\right)\cup \left[{\displaystyle \frac{3}{2}},\mathrm{\infty }\right)$.

Note: Domain of a function is the set of all the values that the variable can take. Range of a function is the set of all the values that the function gives for the values of variables in the domain. While calculating the range we must consider that in the range of the function $x^2\ne 2$ and change the interval accordingly. To verify our answer, we can check what are the values of y when $x^2$ tends to infinity, 2, zero and negative infinity.