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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)

 Jan 4, 2022
 #1
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Answer: 6

Explanation:

The only reason I can think of for there being a number not in the domain of the function is because it causes a denominator to equal zero.

 

For 1/(x-64), there is only one number that causes the denominator to equal 0; namely, 64.

For 1/(x²-64), there are two numbers that causes the denominator to equal 0, namely 8 and -8.

For 1/(x³-64), there is only one number that causes the denominator to equal 0, namely the cube root of 4.

For 1/(x⁴-64), there are two numbers that causes the denominator to equal 0, namely the square root of 8 and negative square root of 8.

 

That is six real numbers.

 

Another way to do it would be to go onto Desmos and count the number of vertical asymptotes of the function (6).

 

(Desmos graph: https://www.desmos.com/calculator/fphhuwje2b)

 

 Jan 4, 2022
edited by WhyamIdoingthis  Jan 4, 2022

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