+0  
 
0
88
2
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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)

 Jun 30, 2021
 #1
avatar+795 
0

This happens when $x = 64, x^2 = 64, x^3 = 64, x^4 = 64$ because then the denominator will be $0.$

Thus the answer is $4.$

If you are looking for the values, they are $64, 8, 4, 2.$

 Jun 30, 2021
 #2
avatar+32658 
+1

Don't forget  -8 and  \(\pm 64^{1/4}\) (fourth root of 64 is real, but not an integer). 

 

Also 2 is in the domain of the function.

Alan  Jun 30, 2021
edited by Alan  Jun 30, 2021

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