Inverse trig Inequality

  • #1
123
5

Homework Statement


Solution set of the inequality (cot-1(x))2 -(5 cot-1(x)) +6 >0 is?

Homework Equations




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?
 

Answers and Replies

  • #2
240
42
An idea: I think you have to mind if ##\cot 2>\cot 3## or ##\cot 3>\cot 2.##
 
  • #3
Ray Vickson
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Homework Statement


Solution set of the inequality (cot-1(x))2 -(5 cot-1(x)) +6 >0 is?

Homework Equations




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?
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.
 
  • #4
SammyS
Staff Emeritus
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@takando12 ,

What is the range of the arccotangent function as you are using it in your course?
 
  • #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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