0
$\begingroup$

What is the domain and range of real function $f(x) = \sqrt{9 − x^2}$?

In order to find the function's domain, you need to take into account the fact that, for real numbers, you can only take the square root of a positive number. In other words, in order for the function to be defined, you need the expression that's under the square root to be positive. \begin{align*} 9−x^2 & \geq 0\\ x^2 & \leq 9\\ |x| & \leq 3 \end{align*} This means that you have $x \geq −3$ and $x \leq 3$.

Therefore, the domain of the function will be $x \in [−3,3]$.

But what will be the range?

Help appreciated!

$\endgroup$
  • 3
    $\begingroup$ Have you tried drawing a graph? $\endgroup$ – hmakholm left over Monica Jun 25 '16 at 11:43
  • 3
    $\begingroup$ Hint: $y=\sqrt{9-x^2}\implies x^2+y^2=9$. $\endgroup$ – user170039 Jun 25 '16 at 11:45
  • $\begingroup$ so is this a circle with radius 3 ? $\endgroup$ – Cyril Cherian Jun 25 '16 at 11:47
  • $\begingroup$ Not quite. Notice that $f(x)$ can't be negative. $\endgroup$ – user137731 Jun 25 '16 at 11:47
  • 1
    $\begingroup$ If you're convinced that your teacher defines $\sqrt{}$ as a multivalued relation, then you're right: the range will be $[-3,3]$. But I think that is unlikely. -- Especially since the notation $f(x)$ is almost exclusively used for functions, not relations. $\endgroup$ – user137731 Jun 25 '16 at 12:04
1
$\begingroup$

The usual convention is that the principal square root function $f(x) = \sqrt x$ returns only the non-negative root. So in this case, the range corresponds to the semicircular upper half of the circle defined by $x^2 + y^2 = 9$. Which gives the range as $[0,3]$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ You are saying usual convention... where is that convention written? This statement below is not really correct The usual convention is that the square root function f(x)=x√f(x)=x returns only the non-negative root $\endgroup$ – Cyril Cherian Jun 25 '16 at 11:56
  • $\begingroup$ $y = \sqrt{x}$ means the principal square root. See my response to this question about why $f(x) = \sqrt{x}$ is a function. $\endgroup$ – N. F. Taussig Jun 25 '16 at 11:59
  • 1
    $\begingroup$ @Cyril Added the word "principal" for clarity. $\endgroup$ – Deepak Jun 25 '16 at 12:01
1
$\begingroup$

Hint:

Since a square is non-negative, $0\le 9-x^2\le9\;$ on $\;[-3,3]$, and $\;\sqrt x\;$ is a continuous increasing function.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

$$9-x^2\ge 0\implies D_f=[-3,3]$$ set $x=3\sin\theta$ $$y=\sqrt{9-9\sin^2\theta}=3\,|\,\cos\theta\,|\implies R_f=[0,3]$$

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

You correctly found the domain, although you meant that the expression that's under the square root, which is called the radicand, must be non-negative.

Observe that $y = \sqrt{9 - x^2} \implies y \geq 0$. Moreover, if we square both sides of the equation, we obtain $y^2 = 9 - x^2$, which is equivalent to $$x^2 + y^2 = 9$$ This is the equation of a circle with radius $3$ and center at the origin. The restriction $y \geq 0$ means that we obtain the upper semi-circle, from which you can determine that the range is $[0, 3]$.

upper_semi-circle

Addendum: I gather from the comments that you have left that you were not aware that the notation $y = \sqrt{x}$ means the principal (non-negative) square root of $x$. That is what allows us to conclude that $y \geq 0$.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ appreciate your effort! Unfortunately i could accept only one answer.. Thanks once again! $\endgroup$ – Cyril Cherian Jun 25 '16 at 12:21
1
$\begingroup$

You just need to find the absolute maxima or minima of $f(x)$ in its domain $[-3,3]$. Write $f(x)=√g(x)$, where $g(x)=9-x^2$. Now, $g(x)$ has an extrema if $g'(x)=0$ which gives $x=0$ as the point of maxima as $g''(0)<0$. Hence $f(0)=√g(0)=3$, $f(-3)=f(3)=0$. Clearly, $Range(f)=[0,3]$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.