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amy0799

  • one year ago

consider the following. f(x) = (x)/(x^2-2) Find the x-values at which f is not continuous, smaller value = larger value = Is the discontinuity at the smaller value removable? Is the discontinuity at the larger value removable?

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  1. anonymous
    • one year ago
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    look for a value of x that makes the denominator zero, it won't be continuous there

  2. Michele_Laino
    • one year ago
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    hint: for continuity, we can not divide by zero

  3. amy0799
    • one year ago
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    i'm confused on what to do...

  4. Michele_Laino
    • one year ago
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    we have tosolve this equation, first: \[\Large {x^2} - 2 = 0\]

  5. amy0799
    • one year ago
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    x=sqrt(2)

  6. Michele_Laino
    • one year ago
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    more precisely, we have two solutions: \[\Large x = \pm \sqrt 2 \]

  7. Michele_Laino
    • one year ago
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    am I right?

  8. amy0799
    • one year ago
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    yea. so the smaller value is \[-\sqrt{2}\] and the larger value is \[\sqrt{2} \]?

  9. Michele_Laino
    • one year ago
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    Correct! we have to exclude both values, since we can not divide by zero, so our function is not continuous at these points: \[\Large \begin{gathered} x = \sqrt 2 \hfill \\ x = - \sqrt 2 \hfill \\ \end{gathered} \]

  10. amy0799
    • one year ago
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    right. so how do you find out if it's removable?

  11. Michele_Laino
    • one year ago
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    I think that both discontinuities are not removable

  12. amy0799
    • one year ago
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    can you explain why?

  13. Michele_Laino
    • one year ago
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    yes! I explain with an example

  14. Michele_Laino
    • one year ago
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    for example, the subsequent function: \[\Large f\left( x \right) = \frac{{{x^2} - 1}}{{x - 1}}\] is not continuous at x=1, right?

  15. amy0799
    • one year ago
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    yes

  16. Michele_Laino
    • one year ago
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    nevertheless x=1 is a removable discontinuity, since we can write this: \[\Large f\left( x \right) = \frac{{{x^2} - 1}}{{x - 1}} = \frac{{\left( {x - 1} \right)\left( {x + 1} \right)}}{{x - 1}} = x + 1\]

  17. amy0799
    • one year ago
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    ooh ok. can you help me with another problem?

  18. Michele_Laino
    • one year ago
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    yes!

  19. amy0799
    • one year ago
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    consider the following. f(x) = {-9x

  20. Michele_Laino
    • one year ago
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    ok!

  21. amy0799
    • one year ago
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    f(x) = -9 if x less than or equal to 5 x^2-2x+2 if x greater than 5

  22. amy0799
    • one year ago
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    find the x-value at which f is not continuous and is the discontinuity removable

  23. Michele_Laino
    • one year ago
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    at x= 5, we have f(x)= 9 using your first part for f(x) and we have: \[{x^2} - 2x + 2 = 25 - 2 \cdot 5 + 2 = 17 \ne 9\] so the discontinuity is at x=5

  24. amy0799
    • one year ago
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    is it not removable?

  25. Michele_Laino
    • one year ago
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    I think that it is not removable

  26. Michele_Laino
    • one year ago
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    because it is a jump-type discontinuity

  27. amy0799
    • one year ago
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    ok thanks!. I have a few more I need help me, is that ok?

  28. Michele_Laino
    • one year ago
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    ok!

  29. amy0799
    • one year ago
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    g(x) = 5sin(x)/x if x less than 0 a-4x if x greater than or less than 0 Find the constant a such that the function is continuous on the entire real line.

  30. Michele_Laino
    • one year ago
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    we can write this: \[\Large \mathop {\lim }\limits_{x \to 0} \frac{{5\sin x}}{x} = 5\]

  31. Michele_Laino
    • one year ago
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    whereas at x=0, we have: \[\Large a - 4x = a - 4 \cdot 0 = a\]

  32. Michele_Laino
    • one year ago
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    now, for continuity, we have to request taht both values are equal each to other. So what do you get?

  33. amy0799
    • one year ago
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    Would the constant a be 5?

  34. Michele_Laino
    • one year ago
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    that*

  35. Michele_Laino
    • one year ago
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    correct! a=5

  36. amy0799
    • one year ago
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    g(x) = (x^2-a^2)/(x-a) if x is not equal to a 11 if x=a Find the constant a such that the function is continuous on the entire real line.

  37. Michele_Laino
    • one year ago
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    we can write this: \[\Large \frac{{{x^2} - {a^2}}}{{x - a}} = \frac{{\left( {x - a} \right)\left( {x + a} \right)}}{{x - a}} = x + a\]

  38. Michele_Laino
    • one year ago
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    and its value at x=a, is: \[\Large 2a\] Now, for continuity we have to request, as before, that those two values are equal each to other, so we have to solve this equation: \[\Large 2a = 11\]

  39. amy0799
    • one year ago
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    I don't understand how you got 2a

  40. Michele_Laino
    • one year ago
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    after a simplification, I got this: \[\Large g\left( x \right) = x + a\] so, \[\Large g\left( a \right) = a + a = 2a\]

  41. amy0799
    • one year ago
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    oh ok. so the constant a is 11/2?

  42. Michele_Laino
    • one year ago
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    yes! correct!

  43. amy0799
    • one year ago
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    Describe the interval(s) on which the function below is continuous. \[f(x)=x \sqrt{x+1}\]

  44. Michele_Laino
    • one year ago
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    your function is continuous inside all its domain, and its domain is the subset of all real numbers, such that the radicand is positive or equal to zero, namely: \[\Large x + 1 \geqslant 0\]

  45. amy0799
    • one year ago
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    [0,∞) is that how you would write it?

  46. Michele_Laino
    • one year ago
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    we have: \[\Large x + 1 \geqslant 0 \Rightarrow x \geqslant - 1\] so, the requested interval is: \[\Large [ - 1,\infty )\] am I right?

  47. amy0799
    • one year ago
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    yes. Thank you so much for the help and sorry if I took up your time.

  48. Michele_Laino
    • one year ago
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    no worries! :) :)

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