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Implied Dom/Ran = tan(2arcsin(x))

  1. Aug 6, 2011 #1
    1. The problem statement, all variables and given/known data

    State the implied domain and range of y= tan(2arcsin(x))

    3. The attempt at a solution

    Let g : [-1,1] [itex]\rightarrow[/itex] [-[itex]\pi[/itex],[itex]\pi[/itex]] , be the function g(x) = 2arcsin(x)

    Let f : (-[itex]\frac{\pi}{2}[/itex],[itex]\frac{\pi}{2}[/itex]) [itex]\rightarrow[/itex] R, be the function f(x) = tan(x)

    So, (f[itex]\circ[/itex]g)(x) = tan(2arcsin(x))

    ran(g) [itex]\cap[/itex] dom(f) = [-[itex]\pi[/itex],[itex]\pi[/itex]] [itex]\cap[/itex] (-[itex]\frac{\pi}{2}[/itex],[itex]\frac{\pi}{2}[/itex]) = (-[itex]\frac{\pi}{2}[/itex],[itex]\frac{\pi}{2}[/itex])

    Restricting dom(g) so that ran(g) = (-[itex]\frac{\pi}{2}[/itex],[itex]\frac{\pi}{2}[/itex]),

    = -[itex]\frac{\pi}{2}[/itex] < 2arcsin(x) < [itex]\frac{\pi}{2}[/itex]

    = -[itex]\frac{\pi}{4}[/itex] < arcsin(x) < [itex]\frac{\pi}{4}[/itex]

    = -[itex]\frac{1}{\sqrt{2}}[/itex] < x < [itex]\frac{1}{\sqrt{2}}[/itex]

    Hence, ran(f[itex]\circ[/itex]g)(x) = R

    Hence, dom(f[itex]\circ[/itex]g)(x) = [-[itex]\frac{1}{\sqrt{2}}[/itex],[itex]\frac{1}{\sqrt{2}}[/itex]]
    But the above answer (implied domain of f(g(x))) is wrong.

    Note : The answer given to me is [-1,1]\{[itex]\pm[/itex][itex]\frac{1}{\sqrt{2}}[/itex]}. It kind of scares me how far off my answer is...

    Using the above method (I draw a visual aid to help me...), I seemed to have no problems finding implied dom/ran for other equations such as y = arcsin(1 - x), y = arccos(2x + 3), y = arctan(4 - x), y = arccos(sin(2x)), etc.

    Any help will definitely be appreciated ^_^... I've been pondering about this question for an hour+...
    Calculus really scares me sometimes =(
     
    Last edited: Aug 6, 2011
  2. jcsd
  3. Aug 6, 2011 #2

    vela

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    You got the domain for f wrong. For example, tan x is defined at [itex]x=\pi[/itex], right? So the domain of f can't be only [itex](-\frac{\pi}{2},\frac{\pi}{2})[/itex].
     
  4. Aug 6, 2011 #3
    Thanks for that =). I'm always forgetting to show the full domain in my trig functions...
    But, it seems that I'm still a little stuck with the implied domain =(

    1. The problem statement, all variables and given/known data

    State the implied domain and range of y= tan(2arcsin(x))

    3. The attempt at a solution

    Let g : [-1,1] [itex]\rightarrow[/itex] [-[itex]\pi[/itex],[itex]\pi[/itex]] , be the function g(x) = 2arcsin(x)

    Let f : (-[itex]\frac{\pi}{2}[/itex] + k[itex]\pi[/itex],[itex]\frac{\pi}{2}[/itex] + k[itex]\pi[/itex]), where k [itex]\in[/itex] Z [itex]\rightarrow[/itex] R, be the function f(x) = tan(x)

    So, (f[itex]\circ[/itex]g)(x) = tan(2arcsin(x))

    ran(g) [itex]\cap[/itex] dom(f) = [-[itex]\pi[/itex],[itex]\pi[/itex]] [itex]\cap[/itex] (-[itex]\frac{\pi}{2}[/itex] + k[itex]\pi[/itex],[itex]\frac{\pi}{2}[/itex] + k[itex]\pi[/itex]), where k [itex]\in[/itex] Z = (-[itex]\frac{\pi}{2}[/itex],[itex]\frac{\pi}{2}[/itex])

    Restricting dom(g) so that ran(g) = (-[itex]\frac{\pi}{2}[/itex],[itex]\frac{\pi}{2}[/itex]),

    = -[itex]\frac{\pi}{2}[/itex] < 2arcsin(x) < [itex]\frac{\pi}{2}[/itex]

    = -[itex]\frac{\pi}{4}[/itex] < arcsin(x) < [itex]\frac{\pi}{4}[/itex]

    = -[itex]\frac{1}{\sqrt{2}}[/itex] < x < [itex]\frac{1}{\sqrt{2}}[/itex]

    Hence, ran(f[itex]\circ[/itex]g)(x) = R

    Hence, dom(f[itex]\circ[/itex]g)(x) = [-[itex]\frac{1}{\sqrt{2}}[/itex],[itex]\frac{1}{\sqrt{2}}[/itex]]

    But implied domain is wrong... Answer given = [-1,1]\{[itex]\pm[/itex][itex]\frac{1}{\sqrt{2}}[/itex]}
     
    Last edited: Aug 6, 2011
  5. Aug 6, 2011 #4

    vela

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    Your domain for f is still wrong. Tangent is defined for all real numbers except odd multiples of π/2. Just picture a number line with holes at ±π/2, ±/2, .... What do you get when you take the intersection of that an [-π/2,π/2]?
     
  6. Aug 6, 2011 #5
    Fingers crossed...
    f : R \ {k[itex]\pi[/itex] + [itex]\frac{\pi}{2}[/itex]}, where k [itex]\in[/itex] Z [itex]\rightarrow[/itex] R, f(x) = tan(x)?

    1. The problem statement, all variables and given/known data

    State the implied domain and range of y= tan(2arcsin(x))

    3. The attempt at a solution

    Let g : [-1,1] [itex]\rightarrow[/itex] [-[itex]\pi[/itex],[itex]\pi[/itex]] , be the function g(x) = 2arcsin(x)

    Let f : R \ {k[itex]\pi[/itex] + [itex]\frac{\pi}{2}[/itex]}, where k [itex]\in[/itex] Z [itex]\rightarrow[/itex] R, be the function f(x) = tan(x)

    So, (f[itex]\circ[/itex]g)(x) = tan(2arcsin(x))

    ran(g) [itex]\cap[/itex] dom(f) = [-[itex]\pi[/itex],[itex]\pi[/itex]] [itex]\cap[/itex] R \ {k[itex]\pi[/itex] + [itex]\frac{\pi}{2}[/itex]}, where k [itex]\in[/itex] Z = (-[itex]\frac{\pi}{2}[/itex],[itex]\frac{\pi}{2}[/itex])

    Restricting dom(g) so that ran(g) = (-[itex]\frac{\pi}{2}[/itex],[itex]\frac{\pi}{2}[/itex]),

    = -[itex]\frac{\pi}{2}[/itex] < 2arcsin(x) < [itex]\frac{\pi}{2}[/itex]

    = -[itex]\frac{\pi}{4}[/itex] < arcsin(x) < [itex]\frac{\pi}{4}[/itex]

    = -[itex]\frac{1}{\sqrt{2}}[/itex] < x < [itex]\frac{1}{\sqrt{2}}[/itex]

    Hence, ran(f[itex]\circ[/itex]g)(x) = R

    Hence, dom(f[itex]\circ[/itex]g)(x) = [-[itex]\frac{1}{\sqrt{2}}[/itex],[itex]\frac{1}{\sqrt{2}}[/itex]]

    But implied domain is wrong... Answer given = [-1,1]\{[itex]\pm[/itex][itex]\frac{1}{\sqrt{2}}[/itex]}

    EDIT : My apologies.. made a number of errors
     
    Last edited: Aug 6, 2011
  7. Aug 6, 2011 #6

    vela

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    Why are you restricting the domain of g this way? The intersection of [itex][-\pi,\pi][/itex] and the domain of tan x isn't [itex](-\pi/2,\pi/2)[/itex].
     
  8. Aug 6, 2011 #7
    Ah! Now I'm seeing (to some degree at least), how neglecting the full domain of tan(x) is leading me to my wrong answer.. I hope this attempt is correct now...

    1. The problem statement, all variables and given/known data

    State the implied domain and range of y= tan(2arcsin(x))

    3. The attempt at a solution

    Let g : [-1,1] [itex]\rightarrow[/itex] [-[itex]\pi[/itex],[itex]\pi[/itex]] , be the function g(x) = 2arcsin(x)

    Let f : R \ {k[itex]\pi[/itex] + [itex]\frac{\pi}{2}[/itex]}, where k [itex]\in[/itex] Z [itex]\rightarrow[/itex] R, be the function f(x) = tan(x)

    So, (f[itex]\circ[/itex]g)(x) = tan(2arcsin(x))

    ran(g) [itex]\cap[/itex] dom(f) = [-[itex]\pi[/itex],[itex]\pi[/itex]] [itex]\cap[/itex] R \ {k[itex]\pi[/itex] + [itex]\frac{\pi}{2}[/itex]}, where k [itex]\in[/itex] Z = [-[itex]\pi[/itex],[itex]\pi[/itex]] \ {[itex]\pm[/itex][itex]\frac{\pi}{2}[/itex]}

    Restricting dom(g) so that ran(g) = [-[itex]\pi[/itex],[itex]\pi[/itex]] \ {[itex]\pm[/itex][itex]\frac{\pi}{2}[/itex]},

    = 2arcsin(x) [itex]\neq[/itex] [itex]\pm[/itex][itex]\frac{\pi}{2}[/itex]

    = arcsin(x) [itex]\neq[/itex] [itex]\pm[/itex][itex]\frac{\pi}{4}[/itex]

    = x [itex]\neq[/itex] [itex]\frac{1}{\sqrt{2}}[/itex]

    Hence, ran(f[itex]\circ[/itex]g)(x) = R

    Hence, dom(f[itex]\circ[/itex]g)(x) = [-1,1]\{[itex]\pm[/itex][itex]\frac{1}{\sqrt{2}}[/itex]}

    And it matches the given answer of = [-1,1]\{[itex]\pm[/itex][itex]\frac{1}{\sqrt{2}}[/itex]}

    Thanks A LOT for your help... Umm, unless i'm still not done yet (my working has errors)...
     
  9. Aug 6, 2011 #8

    vela

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    Looks good!
     
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