Domain and range

((x^2)-x))/(x-1)

Domain cannot equal 1, am I right in saying the range is all real numbers except 1???

(edited 11 years ago)

Could you rewrite the expression? It's a bit unclear.

Is it

x^{2}-\frac{x}{x-1}

?

If it is then the domain is

x \neq 1

Range is all real numbers (including 1)

(edited 11 years ago)

Reply 2

As_Dust_Dances_

Yeah the range would be 1<x<1

(X is greater than or equal to 2) and (X is less than or equal to zero)

(edited 11 years ago)

Reply 3

ghostwalker

Original post by kenny629

x^2-x/x-1

Domain cannot equal 1, am I right in saying the range is all real numbers except 1???

If this is really "undergrad" level, you ought to know to be more precise with your notation. Is that

\dfrac{x^2-x}{x-1}\text{ or }x^2-\dfrac{x}{x-1}\text{ or even }x^2-\dfrac{x}{x}-1\text{ or }\dfrac{x^2-x}{x}-1

(edited 11 years ago)

Reply 4

Jam'

Original post by ghostwalker

If this is really "undergrad" level, you ought to know to be more precise with your notation. Is that

\dfrac{x^2-x}{x-1}\text{ or }x^2-\dfrac{x}{x-1}\text{ or even }x^2-\dfrac{x}{x}-1\text{ or }\dfrac{x^2-x}{x}-1

Agreed. It must be the second one though as the other three simplify to linear or at most quadratic expressions. Haha the only way this is undergrad is if it's at UEL or somewhere **** like that.

(edited 11 years ago)

Reply 5

Jam'

Original post by kenny629

((x^2)-x))/(x-1)

Domain cannot equal 1, am I right in saying the range is all real numbers except 1???

That simplifies to just 'x'. In which case domain is all x and range is all real.

Reply 6

ghostwalker

Original post by Jam'

That simplifies to just 'x'. In which case domain is all x and range is all real.

Although the simplified equation supports a domain of all x, the original equation does not, and the domain of the original equation is

\mathbb{R}\backslash\{1\}

as the OP suspected.

(edited 11 years ago)

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