# Thread: Explain Domain and range of function f(x)

1. ## Explain Domain and range of function f(x)

Determine the range and function defined as y=sqrt (25-x^2)

2. y= $
\sqrt{25-x^2}
$

for domain
25-x^2>=0
x^2<=25
therefor x belongs to [-5,5]
hence domain is [-5,5].
Since its minimum value is 0 and maximum 5 and the function is continuous in its domain so its range is [0,5]
note:I hope you know what is range and what is domain.if you have a doubt,you may ask it.

3. Originally Posted by Ngigi
Determine the range and function defined as y=sqrt (25-x^2)
The graph is the top half of a circle of radius 5 and centre at the origin:

$y = \sqrt{25 - x^2} \Rightarrow y^2 = 25 - x^2 \Rightarrow x^2 + y^2 = 5^2$.

From this graph it should be clear what the domain and range are .....

4. The Domain of a given function is the set of 'input' values (In this case , the $x$ values) for which the function is defined.

The function, $y=\sqrt{(25-x^2)}$, is a composed square root function. The domain is found by solving the inequality $25-x^2 \ge 0$.

$25-x^2 \ge 0$
$-x^2 \ge -25$
$x^2 \le 25$
$x^2\leq 25 \implies \sqrt{x^2}\leq 5$

Therefore the domain will be $[-5, 5]$.
__________________

The Range of a given function is the set of 'output' values (In this case , the $y$ values) for the function.

The function, $y=\sqrt{(25-x^2)}$, is a quadratic hence $\pm x$ values will give the same range. (This is a many to one function)

When $x=0 \implies y=\sqrt{ (25-(0)^2)} = 5$. When $x=\pm 5 \implies y = \sqrt{ (25-(\pm 5)^2)} = 0$.

Therefore the range is $[0,5]$
__________________

Visually, when a graph is drawn, the domain and range can be seen.

5. Originally Posted by Air
[...]
$x^2 \le 25$
$x \le \pm 5$
[...]
There aren't many $x$ such that $x\leq \pm 5$

6. Originally Posted by flyingsquirrel
There aren't many $x$ such that $x\leq \pm 5$
I know, It looks strange. Mathematically, this is the correct method to find the domain but how would I write that line then?

7. Originally Posted by Air
I know, It looks strange. Mathematically, this is the correct method to find the domain but how would I write that line then?
$x^2\leq 25 \implies \sqrt{x^2}\leq 5$ and as $\sqrt{x^2}=|x|$ we have $|x|\leq 5$ which is the same as $-5\leq x \leq 5$.