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UsukiDoll

  • 2 years ago

Let the propositional function C (f,a) mean "The function f is continuous at the point a," and let the propositional function of D(f,a) mean "The function f is differentiable at the point a" Using these symbols together with logical symbols, express the following statements.

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  1. UsukiDoll
    • 2 years ago
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    Neither the tangent function nor the secant function is continuous at pi/2. Either a>0 or the natural logarithm function is not differentiable at a. The absolute value function is continuous at 0, but not differentiable at 0.

  2. UsukiDoll
    • 2 years ago
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    so for the first one I got that the sentence is related to C f(a,) because the functions are tangent and secant. I should write it as ~C(f,a) but is just ~C(f,a)?

  3. UsukiDoll
    • 2 years ago
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    Second one. D(f,a) related. it says that either a>0 or the natural logarithm function is not differential at a. if I didn't have the c(f,a) d(f,a) required I would easily put my P as a>0 and Q that long sentence and that would be P V Q.

  4. UsukiDoll
    • 2 years ago
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    Third one is again D(f,a) related. The absolute value function is continuous at 0, but not differentiable at 0. with my p = absolute value function is continuous at 0 q = not differentiable at 0 this is not an implies or bi-conditional.

  5. UsukiDoll
    • 2 years ago
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    P ^ Q but that would read as The absolute value function is continuous at 0, [and] not differentiable at 0.

  6. UsukiDoll
    • 2 years ago
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    but thing is how to apply the C(f,a) and D(f,a) in this?

  7. ganeshie8
    • 2 years ago
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    For second question :- Either a>0 or the natural logarithm function is not differentiable at a. (a >0) V ~D(ln, a <= 0)

  8. UsukiDoll
    • 2 years ago
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    OH OF COURSE! *facepalm* we have to apply the f a in the C or D

  9. UsukiDoll
    • 2 years ago
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    The absolute value function is continuous at 0, but not differentiable at 0. The ORIGINAL D(f,a) states "The function f is differentiable at the point a" F not differentiable at 0 . . . hmm there's no point a

  10. UsukiDoll
    • 2 years ago
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    the first one for C(f,a) f would be neither tangent nor secant a is continous at pi/2

  11. ganeshie8
    • 2 years ago
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    The absolute value function is continuous at 0, but not differentiable at 0. C(f, 0) ^ ~D(f, 0)

  12. ganeshie8
    • 2 years ago
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    you want to translate the given statements to boolean symbols. thats all right ?

  13. UsukiDoll
    • 2 years ago
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    errr it did say that I have to use those special symbols... if I didn't have to, I can easily see the P and Q 's

  14. ganeshie8
    • 2 years ago
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    wat special symbols ?

  15. UsukiDoll
    • 2 years ago
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    so all of C(f,a) is negated on this.

  16. UsukiDoll
    • 2 years ago
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    Let the propositional function C (f,a) mean "The function f is continuous at the point a," and let the propositional function of D(f,a) mean "The function f is differentiable at the point a" Using these symbols together with logical symbols, express the following statements.

  17. ganeshie8
    • 2 years ago
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    i have read that before

  18. UsukiDoll
    • 2 years ago
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    Neither the tangent function nor the secant function is continuous at pi/2 is purely a negative C (f,a) for f being tangent function nor secant function and a continuous at pi/2

  19. ganeshie8
    • 2 years ago
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    First one : Neither the tangent function nor the secant function is continuous at pi/2 is ~C(tan, pi/2) ^ ~C(sec, pi/2)

  20. UsukiDoll
    • 2 years ago
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    thought so...so I have to make it into detail as everything counts.

  21. ganeshie8
    • 2 years ago
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    just convert the statemetns, whats big deal ha

  22. ganeshie8
    • 2 years ago
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    unless we both are not in same page... :o

  23. UsukiDoll
    • 2 years ago
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    well I quickly saw the first one as a double negative. no no it did say to use C f,a and D f,a Which I sort of seen. except the last one was a tad tricky

  24. ganeshie8
    • 2 years ago
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    for me, first and last are easy. middle one is tricky as we need to think a bit

  25. UsukiDoll
    • 2 years ago
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    k new practice question.

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