Question

Find the domain and range of the following real function :
(i) $f(x) = - \left| x \right|$
(ii) $f(x) = \sqrt {9 - {x^2}}$

Answer 1

As the domain is the set of values for which the function is define and the range is the set of that values which we will get after putting the domain values in it , as for part (i) the modules function is define the all values of the $x$ it is input the value came after putting domain that is the range of the function similarly for part (ii)

Complete step-by-step answer:
The domain of function is the set of all values for which the function is define and the range is the set of that values which we will get after putting the domain values in it ,
So in the part (i) $f(x) = - \left| x \right|$
As the modules function is define the all values of the $x$ mean that the function $f(x) = - \left| x \right|$ whatever the value we put in function , the function is define
So the domain of the function $f(x) = - \left| x \right|$ is $\left( { - \infty ,\infty } \right)$
As we know that the modules function always give the positive value mean that
$\left| x \right| \geqslant 0\forall {\text{ , }}x \in \left( { - \infty ,\infty } \right)$
Now multiple by negative sign as when we multiple by negative sign the sign of inequality will change so
$- \left| x \right| \leqslant 0$ or
we know that $f(x) = - \left| x \right|$
Hence $f(x) \leqslant 0$
So the range of the function $f(x) = - \left| x \right|$ is $\left( { - \infty ,0} \right]$
For the part (ii) $f(x) = \sqrt {9 - {x^2}}$
As this function doesn't define at the some points for example , if we put $x = 5$ in the $f(x) = \sqrt {9 - {x^2}}$then the value of function came as imaginary ,
Hence for the value under a root cannot be negative or else the solution is imaginary
Therefore , $9 - {x^2} \geqslant 0$
So ${x^2} \leqslant 9$ implies that the $x \leqslant 3$ or $x \geqslant - 3$ , or we can also write it as ,
$x \in \left[ { - 3,3} \right]$
Hence domain of the function is $\left[ { - 3,3} \right]$
As if we put the domain value in $f(x) = \sqrt {9 - {x^2}}$
The smallest value we get is $0$ when we put the value $x = - 3,3$
And the largest value we get is $3$ when we put $x = 0$
Hence the range of the function $f(x) = \sqrt {9 - {x^2}}$ is $\left[ {0,3} \right]$.

Note: In this question we also find the domain of the function be drawing the graph of the given function e.g $f(x) = - \left| x \right|$ if we draw graph as $y = - \left| x \right|$ draw the graph of it after that the range of x-axis gives the domain while the range of y-axis gives the Range of the function .