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Math Help - range and domain of a function

  1. #1
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    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

    my answer...

    (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....
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  2. #2
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    Quote Originally Posted by bobey View Post
    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

    my answer...

    (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....
    Your iii) seems correct.

    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).
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  3. #3
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    Quote Originally Posted by Prove It View Post
    Your iii) seems correct.

    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...
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  4. #4
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    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.
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