Domain of a Function

Mar 2012
588
32
How to find the domain of the function.

3/(1 +(4/x^2))^(1/2)

when I set the denominator equal to zero, I get stuck

1 + (4/x^2) = 0
x^2 + 4 = 0
x^2 = -4
now cannot take the root of negative
 

romsek

MHF Helper
Nov 2013
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3,079
California
what happens when you set $x=0$ ?
 
Mar 2012
588
32
the denominator becomes (1 + 4/0)^(1/2)

which is undefined
 

romsek

MHF Helper
Nov 2013
6,836
3,079
California
the denominator becomes 1 + 4/0

which is undefined
exactly, so $0$ cannot be in the domain.

Do you see any other numbers which cannot be in the domain?
 
Mar 2012
588
32
negative values will be positive, so I don't see any other numbers

but is it possible to multiply numerator and denominator by x like this?

3x/x(1 +(4/x^2))^(1/2)

then I enter the x inside the root x^2 which will lead to


3x/(x^2 + 4)^(1/2)

then I will have no problem when the x = 0
 

romsek

MHF Helper
Nov 2013
6,836
3,079
California
negative values will be positive, so I don't see any other numbers

but is it possible to multiply numerator and denominator by x like this?

3x/x(1 +(4/x^2))^(1/2)

then I enter the x inside the root x^2 which will lead to


3x/(x^2 + 4)^(1/2)

then I will have no problem when the x = 0
if $x=0$ can you multiply by $\dfrac 1 x$

is $\dfrac 1 x$ even defined?
 
Mar 2012
588
32
I could not and it was not defined before, but now I did some algebra which helped me get rid of 1/x^2
 

romsek

MHF Helper
Nov 2013
6,836
3,079
California
I could not and it was not defined before, but now I did some algebra which helped me get rid of 1/x^2
yeah but that algebra you did involved dividing $x$ which is undefined when $x=0$
 

romsek

MHF Helper
Nov 2013
6,836
3,079
California
when x = 0

now

0/(4)^(1/2) = 0
you start off with

$\dfrac{3}{\sqrt{1+\frac{4}{x^2}}}=\dfrac{3}{\sqrt{ \dfrac{4+x^2}{x^2}}}=\dfrac{3}{\dfrac{\sqrt{4+x^2}}{|x|}}$

now you cleverly multiply by 1 in the form of $\dfrac {|x|}{|x|}$ to obtain $\dfrac{3|x|}{\sqrt{4+x^2}}$

but when $x=0$ the expression $\dfrac {|x|}{|x|}$ is undefined and thus your bit of cleverness is not allowed.