domain and range of inverse trig functions

**differentiate** · October 14th 2009, 02:47 AM

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

Find the domain and range of:

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

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

so for $y = 2tan^{-1}x$ 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) $y=2cos^{-1}x - 1$

I don't understand this one at all

THANKS IN ADVANCE

**mr fantastic** · October 14th 2009, 03:31 AM

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

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

so for $y = 2tan^{-1}x$ 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 ${\color{red}2 \left(-\frac{\pi}{2}\right) < x < 2 \left(\frac{\pi}{2}\right) \Rightarrow -\pi < x < \pi}$ .

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

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

I don't understand this one at all

THANKS IN ADVANCE

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

**josipive** · October 14th 2009, 03:47 AM

you need to find domain of:

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

if we have any function ( $f$ ) that have inverse function ( $f^{-1}$ ) that means that $f$ 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:

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

if I know this:

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

now i know that:

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

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

also I know:

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

so from:

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

now i know that

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

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

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

try now b)

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