The function is discontinuous at x = 0, x = 1/4 and x = -1/4.
To create the extended function, add f(x) = 0.
What does it mean when a function is discontinuous?
A function is discontinuous when the f(x) cannot be defined.
What is an extended function?
An extended function aims to make the function continuous at some values of the variable.
It is given that:
[tex]f(x)=\frac{2x^{2}+16x}{16x^{3}-x}[/tex]
Steps
The function becomes discontinuous when the denominator becomes zero.
∴ 16x³ - x = 0
⇒ x(16x² - 1) = 0
⇒ x = 0 and 16x² -1 = 0
This is further simplified:
16x² -1 = 0
x² = 1/16
x = ±1/4
This means that the function is discontinuous at x = 0, x = 1/4 and x = -1/4.
The function can be extended as follows:
[tex]f(x)=\frac{2x^{2}+16x}{16x^{3}-x}\\f(x)= \lim_{x \to 0} \frac{2x^{2}+16x}{16x^{3}-x}\\= \lim_{x \to 0} \frac{x(2x+16)}{x(16x^{2}-1)}\\= \lim_{x \to 0} \frac{2x+16}{16x^{2}-1}\\[/tex]
We do not have to add anything to f(x) to extend it. Hence zero can be added to it.
Therefore, we have found that the function is discontinuous at x = 0, x = 1/4 and x = -1/4, and that we can create the extended function by adding 0.
Learn more about discontinuity here: https://brainly.com/question/1417220
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