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\(f(x) = \sqrt{x+1/x-1} g(x) = \sqrt{x+1} / \sqrt{x-1}\)

 

explain why f and g are different 

Guest Mar 13, 2017

Best Answer 

 #9
avatar
+16

So how would you resolve this problem ?

\(\displaystyle \sqrt{-1}=\sqrt{\frac{1}{-1}}=\frac{\sqrt{1}}{\sqrt{-1}}=\frac{1}{\imath} \\ \displaystyle\sqrt{-1}=\sqrt{\frac{-1}{1}}=\frac{\sqrt{-1}}{\sqrt{1}}=\frac{\imath}{1}\)

Equating and cross multiplying,

\(\displaystyle \imath^{2}=1.\)

At what point has a mistake been made ?

Guest Mar 15, 2017
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14+0 Answers

 #1
avatar+44 
+3

f is different from g because in f, the square root of the result of the expression is taken, where as in g, the root of x+ 1 and x-1 are divided by each other rather than the result. 

Pasplox  Mar 13, 2017
 #2
avatar+25984 
+10

When x>1 f(x) = g(x) and both are real

 

When -1 < x < 1 both f(x) and g(x) are imaginary

 

When x< -1 f(x) is real but g(x) is imaginary

.

Alan  Mar 13, 2017
 #3
avatar+89777 
+10

Alan I have a query??

 

\(f(x) = \sqrt{\frac{x+1}{x-1}}\qquad g(x) =\frac{ \sqrt{x+1} }{ \sqrt{x-1}}\\ if\;\;x=-5\\ f(-5) = \sqrt{\frac{-4}{-6}}\\ f(-5) = \sqrt{\frac{4}{6}}\\ f(-5) =\frac{2}{\sqrt{6}}\qquad \text{Definitely real}\\~\\ g(-5) =\frac{ \sqrt{-4} }{ \sqrt{-6}}\\ g(-5) =\frac{ 2i }{ i\sqrt{6}}\\ \text{Can't I cancel out the i's and have }\frac{2}{\sqrt{6}}???\)

Melody  Mar 14, 2017
 #5
avatar+25984 
+5

Oops!  Yes you can.

Alan  Mar 14, 2017
 #6
avatar+89777 
+5

ok thanks Alan

Now I will look at a value of x between  -1 and 1      frown

 

\(f(x) = \sqrt{\frac{x+1}{x-1}}\qquad g(x) =\frac{ \sqrt{x+1} }{ \sqrt{x-1}}\\ if\;\;x=0\\ f(0) = \sqrt{\frac{1}{-1}}\\ f(0) = \sqrt{-1}\\ f(0) =i \\~\\ g(0) =\frac{ \sqrt{1} }{ \sqrt{-1}}\\ g(0) =\frac{ 1}{i}\\ =\frac{ i}{-1}\\ =-i\\ so\\ f(0)\ne g(0)\)

Melody  Mar 14, 2017
 #7
avatar
0

Every number has two square roots.

\(\displaystyle \sqrt{-1}= \pm\imath\)

If you choose to say that f(0) = i,

then you are making a choice.

Ditto with g(x).

Guest Mar 15, 2017
 #8
avatar+25984 
+5

No.  If \(x^2=-1\) then \(x=\pm \sqrt{-1}\) 

However \(\sqrt{-1} = i\)

Alan  Mar 15, 2017
 #9
avatar
+16
Best Answer

So how would you resolve this problem ?

\(\displaystyle \sqrt{-1}=\sqrt{\frac{1}{-1}}=\frac{\sqrt{1}}{\sqrt{-1}}=\frac{1}{\imath} \\ \displaystyle\sqrt{-1}=\sqrt{\frac{-1}{1}}=\frac{\sqrt{-1}}{\sqrt{1}}=\frac{\imath}{1}\)

Equating and cross multiplying,

\(\displaystyle \imath^{2}=1.\)

At what point has a mistake been made ?

Guest Mar 15, 2017
 #10
avatar+25984 
+5

The mistake occurs in the first line:  \(\sqrt{\frac{1}{-1}}\ne\frac{\sqrt1}{\sqrt{-1}}\)  

 

Your subsequent development shows that equality here leads to an inconsistency.

 

We need to look at this as follows:  \(\sqrt{\frac{1}{-1}}\rightarrow\sqrt{(\frac{1}{-1})}\rightarrow\sqrt{-1}=i\)

.

Alan  Mar 16, 2017
 #12
avatar+89777 
0

Can I have some clarification on that please Alan?

Are you saying that  you cannot take the squareroot of a negative number unless the negative sign is in the numerator?  

 

Because obviously      \(\frac{1}{-1} \quad \text{does equal } \;\;\frac{-1}{1}\)

Melody  Mar 17, 2017
 #13
avatar+25984 
0

It's to do with PEMDAS.

 

\(\sqrt{\frac{1}{-1}}\rightarrow (\frac{1}{-1})^{1/2}\) 

 

Parentheses first, then exponent.

.

Alan  Mar 17, 2017
 #14
avatar+89777 
0

ok thanks Alan but that still means that

 

 

\(\sqrt{-\frac{1}{2}} \;\;\;\text{Must be interpreted as }\;\;\sqrt{\frac{-1}{2}}\)

 

I suppose if you think of it as  \(\sqrt{-0.5}\)

the problem is fixed, but that still means the - sign must be in the numerator. 

 

Mmmm  still thinking. ............

Melody  Mar 17, 2017
 #11
avatar+25984 
+5

f and g differ between -1<x<1:

 

.

Alan  Mar 16, 2017

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