# What is the domain and range of $y = \sqrt{9 − x^2}$?

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!

• Have you tried drawing a graph? – hmakholm left over Monica Jun 25 '16 at 11:43
• Hint: $y=\sqrt{9-x^2}\implies x^2+y^2=9$. – user170039 Jun 25 '16 at 11:45
• so is this a circle with radius 3 ? – Cyril Cherian Jun 25 '16 at 11:47
• Not quite. Notice that $f(x)$ can't be negative. – user137731 Jun 25 '16 at 11:47
• 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. – user137731 Jun 25 '16 at 12:04

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 $[0,3]$.

• 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 – Cyril Cherian Jun 25 '16 at 11:56
• $y = \sqrt{x}$ means the principal square root. See my response to this question about why $f(x) = \sqrt{x}$ is a function. – N. F. Taussig Jun 25 '16 at 11:59
• @Cyril Added the word "principal" for clarity. – Deepak Jun 25 '16 at 12:01

Hint:

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

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

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 $[0, 3]$.

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 $y \geq 0$.

• appreciate your effort! Unfortunately i could accept only one answer.. Thanks once again! – Cyril Cherian Jun 25 '16 at 12:21

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, $Range(f)=[0,3]$.