# Thread: range and domain of a function

1. ## range and domain of a function

let f(x) = x^3 +x^2 -3x-3 and g(x) =1/(x+1)

(i) factorize f(x) completely and find the roots of f(x) = 0

(ii) find gf and its domain

(iii) let h(x) = x^2 -3. is gf = h? justify...

(iv) find the range of gf

(i) by long division i got f(x) = (x+1) (x^2-3)

since f(x) = 0 , the roots are -1, -sqrt(3) and sqrt(3)

(ii) gf = 1/(x+1) x (x+1) (x^2-3) = (x^2-3). thus the domain is R\{-1}

(iii) h(x) does not equal to gf because domain of h(x) is R while domain for gf is R\{-1}

(iv) i dunno. how to get range of gf??

is my answer i to iii correct?? anyone justify....

2. Originally Posted by bobey
let f(x) = x^3 +x^2 -3x-3 and g(x) =1/(x+1)

(i) factorize f(x) completely and find the roots of f(x) = 0

(ii) find gf and its domain

(iii) let h(x) = x^2 -3. is gf = h? justify...

(iv) find the range of gf

(i) by long division i got f(x) = (x+1) (x^2-3)

since f(x) = 0 , the roots are -1, -sqrt(3) and sqrt(3)

(ii) gf = 1/(x+1) x (x+1) (x^2-3) = (x^2-3). thus the domain is R\{-1}

(iii) h(x) does not equal to gf because domain of h(x) is R while domain for gf is R\{-1}

(iv) i dunno. how to get range of gf??

is my answer i to iii correct?? anyone justify....

For iv) You have a quadratic function with a minimum at $(0, -3)$.

Since the domain excludes $x = 1$, that means $y = 1^2 - 3 = -2$ may need to be excluded.

However, if we check $y = -2$ we find

$-2 = x^2 - 3$

$x^2 = 1$

$x = \pm 1$.

Therefore $y = -2$ IS defined at a point.

So the range is $[-3, \infty)$.

3. Originally Posted by Prove It

For iv) You have a quadratic function with a minimum at $(0, -3)$.

Since the domain excludes $x = 1$, that means $y = 1^2 - 3 = -2$ may need to be excluded.

However, if we check $y = -2$ we find

$-2 = x^2 - 3$

$x^2 = 1$

$x = \pm 1$.

Therefore $y = -2$ IS defined at a point.

So the range is $[-3, \infty)$.
i have one question : how to become familiar to find a domain or range of a function... some function makes me confuse... from the above example, the domain is R\{-1}, according to my lecturer in order to find a range we have to refer to the domain... since the function is not defined at x = -1 than how to find the range so that i can write the range of the function is R\{something}. is my concept is correct or .... plz help me...

4. A function, in its simplest sense, is just a set of pairs of numbers with the condition that no two pairs have the same first member.

Of course, if you are given a formula, like f(x)= 3x+ 1, or $y= x^2$, you can calculate the pairs: if x= 1, f(x)= 3(1)+1= 4 so a pair is (1, 4) or if x= 1 $y= 1^2= 1$ so a pair in that function is (1,1).

But, you still have to know what values you can take for x. That means that, to define a function, we would have to give a formula and the possible values for x- the domain.

If we are only given a formula, then we can look at the "maximal domain"- the set of all values of x for which the formula makes sense. But we can also define a different function having the same formula but a different domain.

For example, $f(x)= x^2$ has maximal domain all real numbers since I can square any number. If I saw only the formula $f(x)= x^2$, then, by convention, I would take the maximal domain.

But I could also define $g(x)= x^2$ with domain "all positive real numbers". That would be a different function from f(x). Or I could define $h(x)= x^2$ with domain [0, 1]. Now h(x) is different from either f(x) or g(x).

If you were given only the formula, $f(x)= x^2- 3$, without anything else said, by convention, its domain would be "all real numbers".

But, in fact, you were given $h(x)= \frac{x^3 +x^2 -3x-3}{x+ 1}$ which is equal to $x^2- 3$ as long as x is not -1 and is not defined at x= -1. h(x) is NOT the same as f(x). They have the same "formula" but different domains.