Domain / range of this function

  • Thread starter zeion
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  • #1
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Homework Statement



I don't remember the exact question right now, but it was something like this:

f(x) = sqrt(4 - x^2)

I needed to give the domain / range of it, and also symmetry I believe.


Homework Equations





The Attempt at a Solution



I know this is a half circle after being graphed, but how can I show the domain and range in a math way?

For domain, I know 4 - x^ must >= 0, so I can solve for it that way.
But what about range?
Can I also just isolate x and then get sqrt with y inside and restrict to real numbers?
 

Answers and Replies

  • #2
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Formally, you should say: if "y" is an element in the domains, then there's an x, so that:

[itex]\sqrt{4 - x^{2}}= y[/itex]
That already means that y [itex]\geq[/itex] 0, since the square root always gives non-negative values.
Squaring the equation:
=>
x2 = 4 - y2

This equation only has a solution if the right side is positive (or zero). therefore:
4 - y2 [itex]\geq[/itex] 0
Check out what inequality you get from that.

Of course you need to combine it with y [itex]\geq[/itex] 0 (squaring equations usually leads to extra solutions) with the inequality you got.

A little less formally, but probably valid - you could use the graph of the function. By finding the absolute maximum and minimum of the function in it's closed range [-2,2] and noting that the function gets any values between them, being continuous in that segment - you can find your range.
 
  • #3
467
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since the square root always gives non-negative values.
[tex] \sqrt4 = \pm 2 [/tex]
 
  • #4
202
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[tex] \sqrt4 = \pm 2 [/tex]
The square root function is defined to be the positive root of a number, unlike the operation of taking a root from a number.
 

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