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Math Help - Translation of the cos-1

  1. #1
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    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???
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  2. #2
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    Quote Originally Posted by sanado View Post
    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.
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    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?
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    Quote Originally Posted by sanado View Post
    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}<br />
   - 1 \leqslant 2x + 1 \leqslant 1 \hfill \\<br />
   - 2 \leqslant 2x \leqslant 0 \hfill \\<br />
   - 1 \leqslant x \leqslant 0 \hfill \\ <br />
\end{gathered} <br />
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  5. #5
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    Quote Originally Posted by sanado View Post
    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.
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  6. #6
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    Ok

    Quote Originally Posted by sanado View Post
    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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