Translation of the cos-1

**sanado** · April 9th 2008, 05:43 AM

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???

**mr fantastic** · April 9th 2008, 06:02 AM

Originally Posted by sanado

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???

You require $-1 \leq 2x + 1 \leq 1 \Rightarrow -2 \leq 2x \leq 0 \Rightarrow -1 \leq x \leq 0$ .

**sanado** · April 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?

**Plato** · April 9th 2008, 02:10 PM

Originally Posted by sanado

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} $

**mr fantastic** · April 9th 2008, 03:54 PM

Originally Posted by sanado

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?

Yes. No.

**Mathstud28** · April 9th 2008, 04:25 PM

Originally Posted by sanado

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?

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?

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