# Thread: inverse function and domain and range

1. ## 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.

2. Hi
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 $x=g^{-1}(y)$.

3. Originally Posted by flyingsquirrel
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

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...