# Translation of the cos-1

• Apr 9th 2008, 06:43 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, 07:02 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 $-1 \leq 2x + 1 \leq 1 \Rightarrow -2 \leq 2x \leq 0 \Rightarrow -1 \leq x \leq 0$.
• Apr 9th 2008, 02:42 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, 03:10 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.
$\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, 04:54 PM
mr fantastic
Quote:

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