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Math Help - inverse function and domain and range

  1. #1
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    Question inverse function and domain and range

    Hey guys the sin is just throwing me out here and I am not to confident with inverse functions

    Thanks for any help

    Let f (v) = sin v, and g(v) = 1/(2v−1) .
    (a) Find g ◦ f and state the domain and range.
    (b) Is g(v) one-to-one? Find g−1 (v), the inverse function of g(v), and state the domain and range.
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi
    Quote Originally Posted by smplease View Post
    Let f (v) = sin v, and g(v) = 1/(2v−1) .
    (a) Find g ◦ f and state the domain and range.
    g\circ f(x)=\frac{1}{2\sin x -1} This is a fraction hence it exists iff the denominator does not equal 0 : 2\sin x -1\neq 0 \Longleftrightarrow \sin x \neq \frac{1}{2} \Longleftrightarrow x \neq ?

    To find the range, start from -1\leq \sin x \leq 1 then multiply by 2 : -2\leq 2\sin x \leq 2 then add 1...
    (b) Is g(v) one-to-one? Find g−1 (v), the inverse function of g(v), and state the domain and range.
    g is one-to-one iff g'(x)\neq0 for any x belonging to the domain of g. To find g^{-1}, you can write y=\frac{1}{2x-1} and solve this for x : it'll give x=g^{-1}(y).
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  3. #3
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    Quote Originally Posted by flyingsquirrel View Post
    Hi

    g\circ f(x)=\frac{1}{2\sin x -1} This is a fraction hence it exists iff the denominator does not equal 0 : 2\sin x -1\neq 0 \Longleftrightarrow \sin x \neq \frac{1}{2} \Longleftrightarrow x \neq ?

    To find the range, start from -1\leq \sin x \leq 1 then multiply by 2 : -2\leq 2\sin x \leq 2 then add 1...
    g is one-to-one iff g'(x)\neq0 for any x belonging to the domain of g. To find g^{-1}, you can write y=\frac{1}{2x-1} and solve this for x : it'll give x=g^{-1}(y).
    hey thanks heaps for the help...i'm just still a bit confused with the range. How do I add one to all of that?where do i add it to?

    thanks again
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  4. #4
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by smplease View Post
    hey thanks heaps for the help...i'm just still a bit confused with the range. How do I add one to all of that?where do i add it to?
    Sorry, I meant
    Quote Originally Posted by flyingsquirrel View Post
    To find the range, start from -1\leq \sin x \leq 1 then multiply by 2 : -2\leq 2\sin x \leq 2 then subtract 1...
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