Question
Find the domain and range of the real function: $\begin{equation} f(x)=\sqrt{9-x^{2}} \end{equation}$
✓
Answer
Here the given function is: $\begin{equation} f(x)=\sqrt{9-x^{2}} \end{equation}$
Domain: These are the values of $x$ for which $f(x)$ is defined. From the given $f(x)$ we can say that, $f(x)$ should be real and for that, $9 - x^2 \geq 0 [$since a value less than $0$ will give an imaginary value$] (3 + x)(3 - x) \geq 0.$ Now there are two critical points, $x = +3$ and $x = -3$. Taking a value less than $-3$ and putting in the expression we get, $(3 - 5)(3 + 5) = -ve$ value and thus plotting these on number line we have
Since, $f(x)$ is defined for all real numbers that are greater than or equal to $-3$ and less than or equal to $3,$ the domain of $f(x)$ is $[-3, 3].$
Range: The values of $f(x)$ obtained by putting possible values of $x$. From the $f(x)$ we can see that, the values obtained will only be positive and can be any positive number less than $3$. Hence the range of $f(x) = [0, 3).$
Domain: These are the values of $x$ for which $f(x)$ is defined. From the given $f(x)$ we can say that, $f(x)$ should be real and for that, $9 - x^2 \geq 0 [$since a value less than $0$ will give an imaginary value$] (3 + x)(3 - x) \geq 0.$ Now there are two critical points, $x = +3$ and $x = -3$. Taking a value less than $-3$ and putting in the expression we get, $(3 - 5)(3 + 5) = -ve$ value and thus plotting these on number line we have
Since, $f(x)$ is defined for all real numbers that are greater than or equal to $-3$ and less than or equal to $3,$ the domain of $f(x)$ is $[-3, 3].$
Range: The values of $f(x)$ obtained by putting possible values of $x$. From the $f(x)$ we can see that, the values obtained will only be positive and can be any positive number less than $3$. Hence the range of $f(x) = [0, 3).$
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