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anonymous

  • 5 years ago

Find the dicontinuites of any: 1. f(x) = sin(x^2 - 2) 2. f(x) = 1/(1 - 2sinx)

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  1. anonymous
    • 5 years ago
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    can you please describe why because i don't understand discontinuites

  2. anonymous
    • 5 years ago
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    discontinuites*

  3. anonymous
    • 5 years ago
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    Discontinuities happen when the function is undefined for some reason. The biggest ones are when the denominator is zero, or you have a negative number under a square root. Since sine is defined for ANY number you plug in, there are no discontinuities for the first one. For the second one, we have an expression involving x in the denominator, so that should throw up a red flag. The function will be discontinuous when the denominator is 0 (since this would make the function go to infinity, which is undesirable...), and this happens when \[1-2\sin(x) = 0\], or when \[\sin(x) = \frac{1}{2}\] Now just solve for x, and you'll be done!

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