# Translation of the cos-1

• Apr 9th 2008, 05:43 AM
Translation of the cos-1
Hey guys, was just looking through some inverse questions and i got stuck at a certain one. Upon looking at the answers i was shocked, i had never seen anything like it.

3Cos-1(2x+1) - pie/2, Find the domain.

Okay so the only thing that affects the domain in this function is the 2x+1. Of the cos-1(x) function, the domain is -1,1. So since 2x = -1, x= -1/2, shouldnt the new domain be [-1.5,.5]. The answer they gave me was [-1,0]. Is their some rule that applies for shifting the inverse functions across???(Worried)
• Apr 9th 2008, 06:02 AM
mr fantastic
Quote:

Hey guys, was just looking through some inverse questions and i got stuck at a certain one. Upon looking at the answers i was shocked, i had never seen anything like it.

3Cos-1(2x+1) - pie/2, Find the domain.

Okay so the only thing that affects the domain in this function is the 2x+1. Of the cos-1(x) function, the domain is -1,1. So since 2x = -1, x= -1/2, shouldnt the new domain be [-1.5,.5]. The answer they gave me was [-1,0]. Is their some rule that applies for shifting the inverse functions across???(Worried)

You require $\displaystyle -1 \leq 2x + 1 \leq 1 \Rightarrow -2 \leq 2x \leq 0 \Rightarrow -1 \leq x \leq 0$.
• Apr 9th 2008, 01:42 PM
Okay, so if i was given a question like this for a sin-1, it would be the same process? Does this method only apply to the inverse of the trig functions?
• Apr 9th 2008, 02:10 PM
Plato
Quote:

3Cos-1(2x+1) - pi/2, Find the domain.
The answer they gave me was [-1,0].

I really don't understand your problem.
$\displaystyle \begin{gathered} - 1 \leqslant 2x + 1 \leqslant 1 \hfill \\ - 2 \leqslant 2x \leqslant 0 \hfill \\ - 1 \leqslant x \leqslant 0 \hfill \\ \end{gathered}$
• Apr 9th 2008, 03:54 PM
mr fantastic
Quote:

another good example is this...$\displaystyle \ln(u(x))$...since $\displaystyle u(x)>0$ for the function to be defined we find where $\displaystyle u(x)$ is positive....so lets say we had $\displaystyle \ln(x^2+3x-4)$...so$\displaystyle x^2+3x-4>0$...so I do it this way $\displaystyle (x+4)(x-1)=0,x=-4,x=1$...so we set up test intervals and test an element of each interval in the function $\displaystyle f(-5)>0$,$\displaystyle f(-2)<0$,$\displaystyle f(2)>0$...we tested those based on our intervals $\displaystyle (-\infty,-4),(-4,1),(1,\infty)$...so $\displaystyle \ln(x^2+3-4)$ is only defined $\displaystyle \forall{x}\in(-\infty,-4)\bigcup{(1,\infty)}$...make sense?