Results 1 to 4 of 4

Thread: inverse function and domain and range

  1. #1
    Junior Member
    Joined
    Mar 2008
    Posts
    55

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
    Posts
    802
    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.
    $\displaystyle g\circ f(x)=\frac{1}{2\sin x -1}$ This is a fraction hence it exists iff the denominator does not equal 0 : $\displaystyle 2\sin x -1\neq 0 \Longleftrightarrow \sin x \neq \frac{1}{2} \Longleftrightarrow x \neq ?$

    To find the range, start from $\displaystyle -1\leq \sin x \leq 1$ then multiply by 2 : $\displaystyle -2\leq 2\sin x \leq 2$ 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.
    $\displaystyle g$ is one-to-one iff $\displaystyle g'(x)\neq0$ for any $\displaystyle x$ belonging to the domain of $\displaystyle g$. To find $\displaystyle g^{-1}$, you can write $\displaystyle y=\frac{1}{2x-1}$ and solve this for $\displaystyle x$ : it'll give $\displaystyle x=g^{-1}(y)$.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2008
    Posts
    55
    Quote Originally Posted by flyingsquirrel View Post
    Hi

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

    To find the range, start from $\displaystyle -1\leq \sin x \leq 1$ then multiply by 2 : $\displaystyle -2\leq 2\sin x \leq 2$ then add 1...
    $\displaystyle g$ is one-to-one iff $\displaystyle g'(x)\neq0$ for any $\displaystyle x$ belonging to the domain of $\displaystyle g$. To find $\displaystyle g^{-1}$, you can write $\displaystyle y=\frac{1}{2x-1}$ and solve this for $\displaystyle x$ : it'll give $\displaystyle x=g^{-1}(y)$.
    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
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
    Posts
    802
    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 $\displaystyle -1\leq \sin x \leq 1$ then multiply by 2 : $\displaystyle -2\leq 2\sin x \leq 2$ then subtract 1...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Domain/Range of Inverse Trig Function
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Jan 10th 2011, 07:12 PM
  2. Find the domain range and inverse
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: Jul 28th 2010, 03:47 PM
  3. Replies: 1
    Last Post: Apr 11th 2010, 01:04 PM
  4. Domain/Range, and Inverse
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Jan 16th 2010, 11:55 PM
  5. Domain and Range of the Inverse Quadratic Function
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Mar 29th 2009, 02:31 PM

Search Tags


/mathhelpforum @mathhelpforum