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Consider the functionsf(x) = \sqrt{\dfrac{x+1}{x-1}}\qquad\qquad\text{and}\qquad\qquad g(x) = \dfrac{\sqrt{x+1}}{\sqrt{x-1}}.Explain why f and g are not the same function.

Guest Nov 6, 2014

Best Answer 

 #2
avatar+92805 
+5

Thanks Gino,

I really like this question - thanks anon.

I have don the graphs to show you.

Both are the same in the 1st quad but that is the whole graph for g(x)

f(x) also has solutions in the 2nd quad

https://www.desmos.com/calculator/k24o4e61dn

Melody  Nov 7, 2014
 #1
avatar+17744 
+5

Both functions will be undefined when x = 1.

Beside this:

The denominator of g(x) is undefinded whenever x < 1. therefore, g(x) will be undefined for any x < 1.

In f(x), when both the numerator and the denominator are both negative, the quotient will be positive, therefore, f(x) will be defined. The only time that f(x) will be undefined will be when only one of the numerator/denominator group is negative. So, f(x) will be undefined only in the section -1 < x ≤ 1.

The functions are not equivalent when x ≤ -1; otherwise, they are.

geno3141  Nov 7, 2014
 #2
avatar+92805 
+5
Best Answer

Thanks Gino,

I really like this question - thanks anon.

I have don the graphs to show you.

Both are the same in the 1st quad but that is the whole graph for g(x)

f(x) also has solutions in the 2nd quad

https://www.desmos.com/calculator/k24o4e61dn

Melody  Nov 7, 2014

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