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Calculating Domain and Range of a function

Posted on 2004-10-26
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I have knowledge about the domail and range of a function but sometimes it is confusing for me to calculate the domain and range of a function(without using any software)

for example
1.   f(x) = (x^2 - 3x + 2)/(x + 1)   , x <> -1
Dom f = R - [-1]   and
Range f = R - (-5 -2*sqrt(6),-5+2*sqrt(6))

2.  g(x) = 6x + 7 when x <= -2
           = 4 - 3x when -2 < x

Dom g = Real Set
Range g = (-ve infinity, 10)

The above are two sample question. i want to understand how calculate the domain and especially range of a function.

Is there any software to just verify the answer.

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Question by:hyynes
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Accepted Solution

by:
BigRat earned 200 total points
ID: 12420401
The two terms come from discrete mathematics, indeed from maps. If x maps to y there is an association for values of x to values of y. One would write  m as a map :-

      m = [x->y]

Ie, a set of mappings. Then the domain of the map is the set of x values and the range the set of y values :-

        Dom(m) = [x],  Range(m) = [y]

So applied to "continuous functions" the domain of the function is the values for which the function exists, the range the values which the function can take.

For f(x) = x, the domian and range are the same, [-oo,+oo]

In f(x) = x² the domain is still [-oo,+oo] (since for all these possible values the function has a result), but the range is [0,+oo] since no negative number may be a square.

In f(x)=sqrt(1-x²) the domain and range are [-1,+1], ie: confined within the unit circle centered on the origin.

So, the best way to calculate domain and range is to draw a graph. This will show (y axis) which range values may be taken for which domian values (x axis).
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Assisted Solution

by:aburr
aburr earned 200 total points
ID: 12425992
For real functions the domain is usually all the reals unless some are excluded explicitly. (In your ex 1 x<> -1.) You cannot calculate the domain. It must be obtained by inspection of the function.
As Big Rat says a graph will help with the range. But if the function becomes large or has a chopped up domain, a little thought helps. Beware of squares and square roots. If you differentiate the function twice and set to zero you will locate all extreme values.  That will give the range for a continuous function. T
There is no software for generally calculating the range but any graphing software will help.

For practice consider
Y = 2*x^2
R = real set
D = o to +infinity
Note the second derivative is zero at x = 0 where y = 0

Consider
Y = 2*x^2   for –2 <= x >= 2
Y = -20        for –2 > x < 2
Here
R = real set
D = 0 to infinity plus –20
Note that the second derivative method works here too (even though it is not continuous)
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by:hyynes
ID: 12449484
how about my example especially 2nd example???
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Expert Comment

by:BigRat
ID: 12471626
For 2) the x in g(x) takes all values of x from -oo to +oo.
The value of g(x) is restricted however to -oo..-5 and +10..-oo. Combining -oo..+10.
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Expert Comment

by:aburr
ID: 12474031
There is no shortcut to finding the range of discontinuous functions. you have to evaluate the functions in the area of the discontinuities. Big Rat's graphs will help.
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