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$\circ$g)(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$\circ$g)(x) = R

Hence, dom(f$\circ$g)(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$\circ$g)(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$\circ$g)(x) = R

Hence, dom(f$\circ$g)(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$\circ$g)(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$\circ$g)(x) = R

Hence, dom(f$\circ$g)(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$\circ$g)(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$\circ$g)(x) = R

Hence, dom(f$\circ$g)(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!