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Sagot :
Answer:
The answer is "[tex]\bold{\frac{2(x+8)}{(4x^2- 1)(4x^2+ 1)}}[/tex]".
Step-by-step explanation:
[tex]\to f(x) = \frac{2x^2+16x}{16x^3-x}[/tex]
[tex]= \frac{2x(x+8)}{x(16x^2-1)}\\\\= \frac{2(x+8)}{(16x^2-1)}\\\\= \frac{2(x+8)}{((4x)^2-(1)^2)}\\\\= \frac{2(x+8)}{(4x^2- 1)(4x^2+ 1)}\\\\[/tex]
Answer: the function is discontinuous at x=0
To create the extended function add f(x)=0,x=0
Step-by-step explanation:
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