Asked by Trish Goal
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)? Thank you!
f(x)=(1/x-64)+(1/x^2-64)+(1/x^3-64)? Thank you!
Answers
Answered by
Reiny
I think you placed some of those brackets in the wrong places, I am sure you meant
f(x) = 1/(x-64) + 1/(x^2 - 64) + 1/(x^3 - 64)
(the way you have them, they serve no purpose)
remember we can't divide by zero, so
x-64 ≠ 0 , ---- x ≠ 64
x^2 - 64 ≠ 0 ----> x ≠ ± 8
x^3 - 64 ≠ 0 ---> x ≠ 4
(the cubic has two more roots, but they are not real)
x ≠ 4, ± 8, 64
f(x) = 1/(x-64) + 1/(x^2 - 64) + 1/(x^3 - 64)
(the way you have them, they serve no purpose)
remember we can't divide by zero, so
x-64 ≠ 0 , ---- x ≠ 64
x^2 - 64 ≠ 0 ----> x ≠ ± 8
x^3 - 64 ≠ 0 ---> x ≠ 4
(the cubic has two more roots, but they are not real)
x ≠ 4, ± 8, 64
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