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Inverse trig Inequality

  1. May 24, 2016 #1
    1. The problem statement, all variables and given/known data
    Solution set of the inequality (cot-1(x))2 -(5 cot-1(x)) +6 >0 is?

    2. Relevant equations


    3. The attempt at a solution
    Subs cot-1(x)=y
    We get a quadratic inequality in y.
    y2-5y+6>0
    (y-2)(y-3)>0
    Using the wavy curve method, the solution set is,
    y∈(-∞,2) ∪(3,∞)
    So cot-1(x)<2 and cot-1(x)>3
    Taking cot on both sides of the inequality,
    x<cot2 and x>cot3
    x∈(-∞,cot2) ∪(cot3,∞)
    Yet the answer is (-∞,cot3)∪(cot2,∞).
    I'm guessing that in the step where I take cot on both sides, I'll have to change the inequality signs as arccot is a decreasing function. Is that where the problem lies?
     
  2. jcsd
  3. May 24, 2016 #2
    An idea: I think you have to mind if ##\cot 2>\cot 3## or ##\cot 3>\cot 2.##
     
  4. May 24, 2016 #3

    Ray Vickson

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    What is the "wavy curve method"?

    To clarify the answer you were given, plot the curve ##y = \text{arccos}(x)## over a broad range of ##x##, such as ##-10 \leq x \leq 10## to see what the regions ##\text{arccos}(x) < 2## and ##\text{arccos}(x) > 3## look like along the ##x##-axis.
     
  5. May 24, 2016 #4

    SammyS

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    @takando12 ,

    What is the range of the arccotangent function as you are using it in your course?
     
  6. May 25, 2016 #5

    ehild

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    upload_2016-5-25_7-8-37.png
    The plot of y=arccotan(x). In what interval of x is y≤2 or y≥3?
     
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