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mathmath333

  • one year ago

Graphical functions

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} &a.) \ f(x)=-f(x) \hspace{.33em}\\~\\ &b.) \ f(x)=f(-x) \hspace{.33em}\\~\\ &c.) \ \normalsize \text{neither even nor odd function} \hspace{.33em}\\~\\ &d.) \ f(x)\ \normalsize \text{doesn't exist at atleast one point of the domain.} \hspace{.33em}\\~\\ \end{align}}\)

  2. mathmath333
    • one year ago
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    |dw:1435416879315:dw|

  3. amoodarya
    • one year ago
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    it is not ? \[f(x)=cotan(x)\] because of vertical retricemptotes ,and roots

  4. mathmath333
    • one year ago
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    i don't have any clue about the equation of the graph

  5. amoodarya
    • one year ago
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    vertical asymptote ! i had a mistake if f(x)=cotan(x) \[f(-x)=\cot(-x)=\frac{\cos(-x)}{\sin(-x)}=\frac{\cos(x)}{-\sin(x)}=-\cot(x) =-f(x)\]

  6. mathmath333
    • one year ago
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    |dw:1435417311725:dw|

  7. amoodarya
    • one year ago
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    it looks : the roots have a symmetry I think they show roots ,nothing more if they are part of figure , this is not a function

  8. amoodarya
    • one year ago
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  9. phi
    • one year ago
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    btw, choice a) f(x)=-f(x) probably should read f(x) = - f(-x) f(x)= -f(x) would mean y=-y which is only true for y=0

  10. mathmath333
    • one year ago
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    m confused is the answer d.)

  11. amoodarya
    • one year ago
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    domain of this function has not 1 point out R in your picture ,\[D_f=\mathbb{R} -\left\{ 0,a,b \right\}\]

  12. mathmath333
    • one year ago
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    what is a and b

  13. alekos
    • one year ago
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    i'd say the lines just indicate zero crossing points and vertical asymptotes

  14. triciaal
    • one year ago
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    From a textbook this is the graph of y = cot x the domain of the secant function is the set of all real numbers except those of the form 1/2 pi + n pi, and the function is continuous on its domain. Because with limit sec x = +/- angle, the graph of the secant function has the lines x = 1/2 pi + nr as vertical asymptotes. There is no intersection of the graph with the x axis because sec x is never zero. The secant function is even because

  15. triciaal
    • one year ago
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    |dw:1435422150442:dw|

  16. mathmath333
    • one year ago
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    in my book answer is d.)

  17. triciaal
    • one year ago
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    and so the graph of the secant function is symmetric with respect to the y axis.

  18. amoodarya
    • one year ago
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    |dw:1435418650098:dw|

  19. amoodarya
    • one year ago
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    I thin\[\pm\pi\]k a, b are

  20. triciaal
    • one year ago
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    I can see D because it says sec x is never zero no intersection with x axis

  21. mathmath333
    • one year ago
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    ok so its d.) , thanks.

  22. mathmath333
    • one year ago
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    |dw:1435419073815:dw| what about this graph

  23. mathmath333
    • one year ago
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    is this even

  24. amoodarya
    • one year ago
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    yes , function you draw is even

  25. mathmath333
    • one year ago
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    |dw:1435419275045:dw|

  26. mathmath333
    • one year ago
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    this graph is even too ?

  27. amoodarya
    • one year ago
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    yes I think you tried to sketch sec(x)

  28. mathmath333
    • one year ago
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    this is even ?

  29. mathmath333
    • one year ago
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    it was not given it is sec x .

  30. amoodarya
    • one year ago
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  31. mathmath333
    • one year ago
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    ok thnx

  32. mathmath333
    • one year ago
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    what is that dotted line btw

  33. amoodarya
    • one year ago
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    vertical asymptote

  34. amoodarya
    • one year ago
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    do you know vertical asymptote ?

  35. mathmath333
    • one year ago
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    yes

  36. mathmath333
    • one year ago
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    not in graph

  37. amoodarya
    • one year ago
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    usually when denominator goes to zero , function go to infinity (usually)

  38. amoodarya
    • one year ago
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    now , whats your choice ?

  39. mathmath333
    • one year ago
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    choice ?

  40. amoodarya
    • one year ago
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    a,b,c,d ?

  41. mathmath333
    • one year ago
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    b.)

  42. mathmath333
    • one year ago
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    i mean it is even function

  43. amoodarya
    • one year ago
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    it is not an even function in your picture I see a odd function

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