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frx

  • one year ago

I have the function \[f(x)=\frac{ \sin(x \pi ) }{ \sin(2x \pi) }\] which is visually presented in the attached file. I'm supposed, only by looking at the graph, to decide * Where in the interval the function is discontinuous? * Where does the curve have a horizontal tangent? * Where does the function have local maximum, local minimum? I can't see these things and would hence appreciate some guidance.

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  1. frx
    • one year ago
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  2. A_clan
    • one year ago
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    (1) By definition, An infinite discontinuity occurs when there is a vertical asymptote at the given x value. So, your x value will be ....

  3. frx
    • one year ago
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    There are multiple x values then, x=-2.5; x=-1.5; x=-0.5; x=0.5; x=1.5;x=2.5

  4. frx
    • one year ago
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    in the interval -3 to 3

  5. A_clan
    • one year ago
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    So, you will have a range of values and not just one value in the solution

  6. A_clan
    • one year ago
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    x={-2.5,-1.5,-0.5,0.5,1.5,2.5}

  7. frx
    • one year ago
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    I get it, thanks! Then how about a horizontal tangent, don't I have to calculate the lim as f(x) goes to infinity or is it possible to see it?

  8. A_clan
    • one year ago
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    It is possible to see it

  9. A_clan
    • one year ago
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    Is there any point on the curve where , on drawing a straight line, you get a horizontal line ?

  10. frx
    • one year ago
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    I can't see it, no. The x-axis itself?

  11. A_clan
    • one year ago
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    |dw:1365268526951:dw| This is one example

  12. frx
    • one year ago
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    Are there two horizontal tangents then? Going through the points of the curvatures?

  13. A_clan
    • one year ago
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    No, every point of curvature has its own tangent

  14. frx
    • one year ago
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    So there are seven horizontal lines which are tangents but the question is asked in singular which should indicate that there where only one answer?

  15. A_clan
    • one year ago
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    My experience is, In mathematics, the grammar of the question may not always indicate the quantity of solution.

  16. frx
    • one year ago
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    So the solutionset is {-3,-2,-1,0,1,2,3} then ?

  17. A_clan
    • one year ago
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    yes, that looks right

  18. frx
    • one year ago
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    Great! What about the local maximum and minimum, there seems to be muliple max and mins too?

  19. A_clan
    • one year ago
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    The solution to last part is situational. Local maximum and minimum may change depending upon the interval you choose for calculating them.

  20. frx
    • one year ago
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    The interval is -3 to 3, like the graph shows

  21. A_clan
    • one year ago
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    We say 'a' is local maximum if the height of the function at 'a' is greater than (or equal to) the height anywhere else in that interval.

  22. A_clan
    • one year ago
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    In simple words, we can say, local maxima is the maximum height in certain part of the graph but if we consider the whole graph ,one of these has to be the 'global maximum'

  23. A_clan
    • one year ago
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    |dw:1365269863585:dw|

  24. A_clan
    • one year ago
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    A,B and C are the local maxima but out of these, B is the global maxima

  25. frx
    • one year ago
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    Sure i understand , so the local maximas are at -3, -1, 1 and 3 and the local mins are -2,0,2

  26. A_clan
    • one year ago
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    That seems right

  27. frx
    • one year ago
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    Should I include the endpoints -3 and 3 according to WA they are not included http://www.wolframalpha.com/input/?i=Local+max+sin%28x+pi%29%2Fsin%282x+pi%29+from+-3+to+3

  28. A_clan
    • one year ago
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    It seems the local maxima and local minima are left out at the boundary because we do not have the values of y from both left side and right side,when x is approaching -3 or 3 .

  29. A_clan
    • one year ago
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    Then correct solution should be -1 and 1

  30. frx
    • one year ago
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    Ok, that makes sense :)

  31. frx
    • one year ago
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    Thanks you so much for your help, really appreciate it!

  32. A_clan
    • one year ago
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    no problem

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