Thread: domain and range of inverse trig functions

1. domain and range of inverse trig functions

hey, can someone tell me how to do these? like explain it and all

Find the domain and range of:

a) $\displaystyle y = 2tan^{-1}x$

do for y = 2tanx.. the domain would be -pi/2 < x < pi/2
the range is all real y

so for $\displaystyle y = 2tan^{-1}x$ to domain should be all real x
and range should be -pi/2 < x < pi/2

But the book said that the range should be -pi< x < pi
I don't get why.

b) $\displaystyle y=2cos^{-1}x - 1$

I don't understand this one at all

2. Originally Posted by differentiate
hey, can someone tell me how to do these? like explain it and all

Find the domain and range of:

a) $\displaystyle y = 2tan^{-1}x$

do for y = 2tanx.. the domain would be -pi/2 < x < pi/2
the range is all real y

so for $\displaystyle y = 2tan^{-1}x$ to domain should be all real x
and range should be -pi/2 < x < pi/2 Mr F says: There's a dilation by a factor of 2 in the vertical direction therefore the range is $\displaystyle {\color{red}2 \left(-\frac{\pi}{2}\right) < x < 2 \left(\frac{\pi}{2}\right) \Rightarrow -\pi < x < \pi}$.

But the book said that the range should be -pi< x < pi
I don't get why.

b) $\displaystyle y=2cos^{-1}x - 1$

I don't understand this one at all

For (b) is it $\displaystyle y=2cos^{-1}(x) - 1$ or $\displaystyle y=2cos^{-1}(x - 1)$ ?

3. you need to find domain of:

$\displaystyle g(x) = 2tan^{-1}(x)$

if we have any function ($\displaystyle f$) that have inverse function ($\displaystyle f^{-1}$) that means that $\displaystyle f$ is bijection ( 1 - 1 )

or:

f: A -> B -----> f^{-1}: B -> A and card( A ) = card( B )

( card ---> cardinal number or number of elements in set )

so if I know that:

$\displaystyle g(x) = 2tan^{-1}(x) = 2arctan(x)$
than it is easy to find $\displaystyle g^{-1}(x)$

if I know this:

$\displaystyle f( x ) = tan(x)$---->$\displaystyle f^{-1}(x) = arctan(x)$

now i know that:

$\displaystyle g(x) = 2 arctan(x)$---->$\displaystyle g^{-1}(x) = \frac{tan(x)}{2}$

and i know that: $\displaystyle R_f = \mathbb{R}$ ($\displaystyle R_f$ is range of tan )

also I know:

$\displaystyle D_f = <\frac{-\pi}{2},\frac{\pi}{2}>$ ($\displaystyle D_f$ is domain of tan )

so from:

f: A -> B -----> f^{-1}: B -> A

now i know that

domain of $\displaystyle 2arctan(x)$ is R,

and i know that range of $\displaystyle arctan(x)$ is $\displaystyle <\frac{-\pi}{2},\frac{\pi}{2}>$

so range of $\displaystyle 2arctan(x)$ is
$\displaystyle <-\pi,\pi>$

try now b)