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

[W0rr13d-0n3]

Homework Statement

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

The Attempt at a Solution

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

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

So, (f\circg)(x) = tan(2arcsin(x))

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

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

= -\frac{\pi}{2} < 2arcsin(x) < \frac{\pi}{2}

= -\frac{\pi}{4} < arcsin(x) < \frac{\pi}{4}

= -\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}

Hence, ran(f\circg)(x) = R

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

Note : The answer given to me is [-1,1]\{\pm\frac{1}{\sqrt{2}}}. 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 =(

 

[vela]

You got the domain for f wrong. For example, tan x is defined at x=\pi, right? So the domain of f can't be only (-\frac{\pi}{2},\frac{\pi}{2}).

 

[W0rr13d-0n3]

You got the domain for f wrong. For example, tan x is defined at x=\pi, right? So the domain of f can't be only (-\frac{\pi}{2},\frac{\pi}{2}).

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 =(

Homework Statement

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

The Attempt at a Solution

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

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

So, (f\circg)(x) = tan(2arcsin(x))

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

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

= -\frac{\pi}{2} < 2arcsin(x) < \frac{\pi}{2}

= -\frac{\pi}{4} < arcsin(x) < \frac{\pi}{4}

= -\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}

Hence, ran(f\circg)(x) = R

Hence, dom(f\circg)(x) = [-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}]

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

 

[vela]

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, ±3π/2, ... What do you get when you take the intersection of that an [-π/2,π/2]?

 

[W0rr13d-0n3]

Fingers crossed...
f : R \ {k\pi + \frac{\pi}{2}}, where k \in Z \rightarrow R, f(x) = tan(x)?

Homework Statement

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

The Attempt at a Solution

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

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

So, (f\circg)(x) = tan(2arcsin(x))

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

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

= -\frac{\pi}{2} < 2arcsin(x) < \frac{\pi}{2}

= -\frac{\pi}{4} < arcsin(x) < \frac{\pi}{4}

= -\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}

Hence, ran(f\circg)(x) = R

Hence, dom(f\circg)(x) = [-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}]

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

EDIT : My apologies.. made a number of errors

 

[vela]

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

Why are you restricting the domain of g this way? The intersection of [-\pi,\pi] and the domain of tan x isn't (-\pi/2,\pi/2).

 

[W0rr13d-0n3]

Why are you restricting the domain of g this way? The intersection of [-\pi,\pi] and the domain of tan x isn't (-\pi/2,\pi/2).

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...

Homework Statement

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

The Attempt at a Solution

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

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

So, (f\circg)(x) = tan(2arcsin(x))

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

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

= 2arcsin(x) \neq \pm\frac{\pi}{2}

= arcsin(x) \neq \pm\frac{\pi}{4}

= x \neq \frac{1}{\sqrt{2}}

Hence, ran(f\circg)(x) = R

Hence, dom(f\circg)(x) = [-1,1]\{\pm\frac{1}{\sqrt{2}}}

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

Thanks A LOT for your help... Umm, unless I'm still not done yet (my working has errors)...

 

[vela]

Looks good!