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anonymous

  • 4 years ago

find the domain and range of f(x)=1+x(square)

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  1. ash2326
    • 4 years ago
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    \[f(x)=1+x^2\] as there is no x for which f(x) is undefined domain= all real numbers the minimum value of f(x)= 1 when x=0 so range = [1, \(\infty\)) or f(x)>=1

  2. perl
    • 4 years ago
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    domain: all reals range : [1 , oo)

  3. perl
    • 4 years ago
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    ash, good explanation ;)

  4. perl
    • 4 years ago
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    we take as the domain the largest possible set of real numbers. since there are no restrictions (no points that are undefined), the domain is all reals

  5. perl
    • 4 years ago
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    , it is assumed that the domain is the largest possible set of reals

  6. ash2326
    • 4 years ago
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    izzati you got it?

  7. anonymous
    • 4 years ago
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    actually i dont understand...it is in exam we should write it like that??

  8. ash2326
    • 4 years ago
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    yeah izzati

  9. anonymous
    • 4 years ago
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    how you got the range..explain plez..

  10. ash2326
    • 4 years ago
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    We have \[f(x)=1+x^2\] we know that x^2 's minimum value is 0 whether x>0 or x<0 , value of square of x will be greater than 0 so its minimum value is 0 so f(x)'s minimum value is 1+0=1 when x^2 increases, f(x) also increases when x -----> infinity , f(x) ----> infinity no maximum limit of f(x) so f(x) 's range is greater than or equal to 1 f(x)>=1 or range is [1, \(\infty\))

  11. anonymous
    • 4 years ago
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    ok.i got it thanx

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