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How many real numbers are not in the domain of the function
f(x) = 1/(x - 64) + 1/(x^2 - 64) + 1/(x^3 - 64) + 1/(x^4 - 64)?

 Apr 15, 2022
 #1
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Whatever x  makes the denominator  = 0  is  not in the  domain

x - 64  =0 

64 is  not in the  domain

 

x^2 - 64  =  0     so.... ( x + 8) ( x -8)    = 0     ....so    x = -8  and  x = 8  are not in the domain

 

x^3 - 64  =  0

x^3 = 64

x = 4    is not in the  domain

 

x^4 - 64   =  0 

(x^2 - 8) (x^2 + 8)   = 0

x^2  - 8  =0  ⇒   x =  ±sqrt (8)

The second factor set to 0  produces a non-real solution 

 

 

So....not in the  domain  =  x =   -8,  -sqrt (8) , 4 , sqrt (8), 8 , 64

 

cool cool cool

 Apr 16, 2022

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