Determine the range and function defined as y=sqrt (25-x^2)
y=$\displaystyle
\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.
The Domain of a given function is the set of 'input' values (In this case , the $\displaystyle x$ values) for which the function is defined.
The function, $\displaystyle y=\sqrt{(25-x^2)}$, is a composed square root function. The domain is found by solving the inequality $\displaystyle 25-x^2 \ge 0$.
$\displaystyle 25-x^2 \ge 0$
$\displaystyle -x^2 \ge -25$
$\displaystyle x^2 \le 25$
$\displaystyle x^2\leq 25 \implies \sqrt{x^2}\leq 5$
Therefore the domain will be $\displaystyle [-5, 5]$.
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The Range of a given function is the set of 'output' values (In this case , the $\displaystyle y$ values) for the function.
The function, $\displaystyle y=\sqrt{(25-x^2)}$, is a quadratic hence $\displaystyle \pm x$ values will give the same range. (This is a many to one function)
When $\displaystyle x=0 \implies y=\sqrt{ (25-(0)^2)} = 5 $. When $\displaystyle x=\pm 5 \implies y = \sqrt{ (25-(\pm 5)^2)} = 0$.
Therefore the range is $\displaystyle [0,5]$
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Visually, when a graph is drawn, the domain and range can be seen.