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Question 17616: please give the range and domain of this function
f(x)=1/{sqrt(x^2-4)}
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! f(x)= 1/{sqrt(4-x^2)}
domain of a function f(x) is set of those values of x which will make the function mathematically legal or correct..certain operations like division by zero , square root of a negative number do not exist in real maths. so they are prohibited .so we have to look carefully for such possibilities to determine the domain of the function..here we have both the operations present
1..division by {sqrt(4-x^2)} which should not equal zero..that is 4-x^2=0 is prohibited..that is 4=x^2 is prohibited..so x=+2 and -2 are prohibited
2...{sqrt(4-x^2)}..square root of (4-x^2)..so 4-x^2 should be zero or positive..zero is already prohibited above ..so it can be only positive ..so
4-x^2>0...or...4>x^2...or ...x^2<4...or...|x|<4^0.5..or |x|<2...or x<2 and...... x>-2......or -2 to sum up the domain is all values of x such that x<2 and x>-2..that is x should lie between -2 and +2.......or -2 Now range of f(x) is the corrsponding values f(x) will take when x takes different values in the specified domain..
we find that as x varies between -2 to 0 and from 0 to +2 ... {sqrt(4-x^2)}varies between infinitely large number to 2 and again from 2 to infinitely large number ..so
f(x)= 1/{sqrt(4-x^2)}varies between 1/(infinitely large number ) to 1/2 that is zero ,but not equal to zero ,to 1/2 so range is f(x)>0 and f(x)<=1/2..
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