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ineedhelp10

  • one year ago

help! question written inside here

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  1. ineedhelp10
    • one year ago
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    Find (f+g)(x),(f-g)(x), (f*g)(x), and (f/g)(x) for each f(x) and g(x). State the domain of each new function. a. f(x)=x^2+4; g(x)=\[\sqrt{x}\]

  2. ineedhelp10
    • one year ago
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    b. f(x)= x-7; g(x)=x+7

  3. ineedhelp10
    • one year ago
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    c. f(x)=\[\sqrt{x+8}; g(x)=\sqrt{x+5}-3\]

  4. ineedhelp10
    • one year ago
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    \[d. f(x)= \frac{ 3 }{ x }; g(x)=x^4\]

  5. ineedhelp10
    • one year ago
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    @timo86m

  6. ineedhelp10
    • one year ago
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    im confused on this :/

  7. anonymous
    • one year ago
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    Find (f+g)(x),(f-g)(x), (f*g)(x), and (f/g)(x) for each f(x) and g(x). State the domain of each new function. (f+g)(x) means add the 2 functions. Also written as f(x)+g(x) (f-g)(x) subtract the 2 functions f(x)-g(x) (f*g)(x) means multiply the 2 functions. Also written as f(x)*g(x) (f/g)(x) means divide f(x) by g(x). f(x)/g(x) Domain means all x values that work. Typical examples 1/x all numbers work except for numbers that make the denominator 0. In this case 0 1/(x+1) Here -1 wont work. SO all numbers except for -1 is domain Another type of domain example is sqrt(x) no numbers less than zero make it not work. So you can say domain is x>=0 >= Means greater or equal to zero

  8. ineedhelp10
    • one year ago
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    so then like for the first one im adding x^2 and \[\sqrt{x}\]

  9. ineedhelp10
    • one year ago
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    ******x^2+4 i mean

  10. anonymous
    • one year ago
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    (x^2+4)+(sqrt(x)) yes I like to use parenthesis at first

  11. ineedhelp10
    • one year ago
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    and what would the answer be? i didnt even know you could add those two together

  12. anonymous
    • one year ago
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    I dont think you can simplify a further so the domain is actually. x>=0

  13. ineedhelp10
    • one year ago
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    so for the addition, you dont do anything right just literally add them together correct?

  14. anonymous
    • one year ago
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    (f-g)(x) same applies here.

  15. anonymous
    • one year ago
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    You are correct. you will add them together if they are like terms like 2x+x=3x x^2+4x^2=5x^2 x+x+x+x=4x

  16. anonymous
    • one year ago
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    The odd pare is the multiplication part. (x^2+4)*(sqrt(x)) (sqrt(x))* (x^2+4) rearrange sqrt(x)*x^2+ 4*sqrt(x) distribute the sqrt (x) again domain will be x>=0 those are the only numbers that make it work. I dont think i even needed to distribute the sqrt(x) here. would still work

  17. ineedhelp10
    • one year ago
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    for the multiplication part i got \[x^2+4\sqrt{x}\]

  18. anonymous
    • one year ago
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    that looks wrong dude. I'll copy paste the right way from above The odd pare is the multiplication part. (x^2+4)*(sqrt(x)) (sqrt(x))* (x^2+4) rearrange sqrt(x)*x^2+ 4*sqrt(x) distribute the sqrt (x) again domain will be x>=0 those are the only numbers that make it work. I dont think i even needed to distribute the sqrt(x) here. would still work

  19. anonymous
    • one year ago
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    anyhoo gotta go

  20. ineedhelp10
    • one year ago
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    kk bye thanks anyway!

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