Thread: [SOLVED] domain and range

Domain and Range are my weakness. I have attempted these problems but I am not sure that they are correct could someone check them/tell me how to correct them please?
Many thanks!
f(x)=1+(x^2) D: all reals R:[1, inf)

f(x)=1-(1/x) D: (-inf, 0) R: (0, inf)

f(x)=sqrt(3x-1) D: (1/3, inf) R: (1/3, inf)

f(x)=(x^3)-8 D: all reals R: all reals

f(x)=x/(x-3) D: (-inf, 3) R: (3, inf)

Also, if i wanted to write the domain as all reals except one do i denote that R-1?

f(x)=x/(x-3) D: (-inf, 3) u (3, inf) R: (-inf, 1) u (1, inf)

Also, if i wanted to write the domain as all reals except one do i denote that R-1?

I think you it's like this: R / {the one that doesnt fit}
e.g. R / {5}
But I'm not sure.

Originally Posted by deathbyproofs

f(x)=1+(x^2) D: all reals R:[1, inf)

yes.

f(x)=1-(1/x) D: (-inf, 0) R: (0, inf)

Nein,

D: (-inf, 0) U (0,inf), R: (-inf, 1) U (1, inf)

f(x)=sqrt(3x-1) D: (1/3, inf) R: (1/3, inf)

Nein, 1/3 is included in the domain. the range is also wrong.

D: [1/3, inf), R: [0,inf)

f(x)=(x^3)-8 D: all reals R: all reals

yes

f(x)=x/(x-3) D: (-inf, 3) R: (3, inf)

Nein,

D: (-inf, 3) U (3, inf), R: (-inf, 1) U (1, inf)

Also, if i wanted to write the domain as all reals except one do i denote that R-1?

you may use this notation, but you would say {1} not just 1. it is a set difference that you are doing. the set of real numbers minus the element of the set {1}

Originally Posted by metlx

I think you it's like this: R / {the one that doesnt fit}
e.g. R / {5}
But I'm not sure.

it is "\" not "/"