Question

How do you find the domain and range of `f(x)=(x+7)/(x^2-49)`?

Answer 1

The domain is `x in RR-{7}`
The range is `f(x) in (-oo,0^-)uu(0^+,+oo)`

Explanation:

Let's simplify the function

`f(x)=(x+7)/(x^2-49)=cancel(x+7)/(cancel(x+7)(x-7))`

`f(x)=1/(x-7)`

The denominator is `!=0`

Therefore,

`x-7!=0`, `=>`,`x!=7`

The domain is `I=RR-{7}`

The function is bijective over the domain `I`

`f(-oo)=1/(-oo-7)=0^-`

`f(7^-)=1/(7^(-) -7)=-oo`

`f(7^+)=1/(7^(+) -7)=+oo`

`f(+oo)=1/(+oo-7)=0^+`

Therefore,

The range is `f(x) in (-oo,0^-)uu(0^+,+oo)`

graph{(1/(x-7)) [-13.34, 27.2, -10.06, 10.22]}