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For what values does the function fail to exist?
f(x)=(x)/(x³-16²)

User Soulshake
by
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1 Answer

7 votes

Final answer:

The function f(x) = x/(x^3 - 256) fails to exist when the denominator equals zero, which is at x = 6.3496. This value is a vertical asymptote of the function.

Step-by-step explanation:

The student is asking about the values for which the function f(x) = \frac{x}{x^3 - 256} fails to exist. This typically occurs at values of x that make the denominator equal to zero, since division by zero is undefined within the realm of real numbers.

First, we need to find when the denominator is zero:

x^3 - 256 = 0
x^3 = 256
x = \sqrt[3]{256}
x = 6.3496 (rounded to 4 decimal places)

Therefore, the function fails to exist when x = 6.3496, which is when the denominator x^3 - 256 equals zero resulting in a division by zero situation.

Understanding when functions are not defined due to asymptotes or limits, as mentioned in the reference information, is crucial in analyzing and graphing functions. The value x = 6.3496 represents a vertical asymptote for the function f(x).

User Wbadry
by
8.4k points
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