Restrictions on Domain

Algebra Level 2

f ( x ) = x 1 x ( 1 x 2 ) f\left( x \right) =\frac { x-1 }{ x\left( 1-{ x }^{ 2 } \right) }

What is the sum of the real values for which the function is undefined?


The answer is 0.

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1 solution

Zeeshan Ali
Jan 23, 2016

f ( x ) = x 1 x ( 1 x 2 ) f\left( x \right) =\frac { x-1 }{ x\left( 1-{ x }^{ 2 } \right) } The function above is not defined at all x for which x ( 1 x 2 ) = 0 x = 0 x\left( 1-{ x }^{ 2 } \right)=0 \implies x=0 or 1 x 2 = 0 x = 0 , 1 , + 1 1-x^2=0\implies x=0,-1,+1 . Hence for x=-1, 0 and +1 the function is not defined. And the required sum is 1 + 0 + 1 = 0 -1+0+1=\boxed 0

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