Find the domain and range of a real function f(x) =sqrt 9- x^2 ?

Question

anonymous · Answer

@satellite73

zepp · Answer

$$\sqrt{9-x^2}$$

anonymous · Answer

here we go again
but this time lets work smarter

zepp · Answer

Do the same thing as @satellite73 showed you :D

anonymous · Answer

start with the equation of a circle with radius 3 and center at the origin. the equation for such a circle is $x^2+y^2=9$ 
and i hope you have seen this before
so lets solve for $y$ and see what happens

anonymous · Answer

wat confused..... do this by that way

anonymous · Answer

$$x^2+y^2=9$$
$$y^2=9-x^2$$
$$y=\pm\sqrt{9-x^2}$$ now this is not a function, because of the $\pm$ but that is ok, get rid of the $-$ part and get $f(x)=\sqrt{9-x^2}$

anonymous · Answer

so what you see is that this function is just the upper half of the circle, i.e. a semicircle, which radius 3
so the domain will be $[-3,3]$ and the range will bet $[0,3]$

anonymous · Answer

@satellite73  any other way!!

anonymous · Answer

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anonymous · Answer

ur answer is correct!

anonymous · Answer

yes, we can do just what we did before

anonymous · Answer

can u show that plzz

experimentx · Answer

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