# Thread: domain and range of a function

1. ## domain and range of a function

Let f(x) = (x+3)/(x+1)

(a) Find the domain and range of f(x)

(b) Show that f(x) is one-to-one function and find the inverse function f-1(x)

My attempt for (a) : since the function is undefined when x = -1 thus the domain is IR\{-1}

But my problem is how i can find the range of the function. is the range of the function is IR except the value when x=-1. BUT how i can find the value of the function at x=-2???

If I were to find the range of the function by graph, how f(x) graph looks like???

next if i want to show that f(x) is one-to-one function then i have to find the inverse of it... is there any other way to show that it is one-to-one???

the way i find the inverse :

let f(x) = y

thus have to find x in terms of y

y = x+3/x+1

but how can i group the x so that i can found x in terms of y??? since they are separated by quotient....

2. Originally Posted by bobey
Let f(x) = (x+3)/(x+1)

(a) Find the domain and range of f(x)

(b) Show that f(x) is one-to-one function and find the inverse function f-1(x)

My attempt for (a) : since the function is undefined when x = -1 thus the domain is IR\{-1}

But my problem is how i can find the range of the function. is the range of the function is IR except the value when x=-1. BUT how i can find the value of the function at x=-2???

If I were to find the range of the function by graph, how f(x) graph looks like???

next if i want to show that f(x) is one-to-one function then i have to find the inverse of it... is there any other way to show that it is one-to-one???

the way i find the inverse :

let f(x) = y

thus have to find x in terms of y

y = x+3/x+1

but how can i group the x so that i can found x in terms of y??? since they are separated by quotient....
Range of f = domain its inverse

to find the inverse:

$\displaystyle y=\frac{x+3}{x+1}$

$\displaystyle yx+y=x+3$

$\displaystyle yx-x=3-y$

$\displaystyle x=\frac{3-y}{y-1}=-\frac{y-3}{y-1}$

so $\displaystyle f^{-1}(x)=-\frac{x-3}{x-1}$

clearly, domain of f inverse is R / {1}

so the range of f is R / {1}