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anonymous

  • one year ago

help please. Find (f + g)(x), (f - g)(x), (f . g)(x), and (f/g)(x) for each f(x) and g(x). 1. f(x) = a^2 - 1 , g(x) = a/(a + 1) (f+g)(x) (f-g)(x) and (f/g)(x)

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  1. freckles
    • one year ago
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    (f+g)(x) means to add f(x) and g(x)

  2. freckles
    • one year ago
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    (f-g)(x) means to take away g(x) from f(x) or just write f(x)-g(x)

  3. freckles
    • one year ago
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    (f/g)(x) means to write f(x)/g(x)

  4. anonymous
    • one year ago
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    ok i have a question

  5. anonymous
    • one year ago
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    what is wrong with this picture \[f(x) = a^2 - 1\]

  6. anonymous
    • one year ago
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    im lost

  7. freckles
    • one year ago
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    it is a constant function satellite :p example: \[f(x)=2 \\ g(x)=3+x \\ \text{ then } (f+g)(x)=2+(3+x)=5+x \\ \text{ also } (f-g)(x)=2-(3+x)=-x-1 \\ \text{ and } (\frac{f}{g})(x)=\frac{2}{3+x}\]

  8. anonymous
    • one year ago
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    me too if you have a function of x, it should have an x in it

  9. anonymous
    • one year ago
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    whats going on?

  10. freckles
    • one year ago
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    well are your functions really f(x)=a^2-1 and g(x)=a/(a+1) ?

  11. anonymous
    • one year ago
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    don't know does it really say \[f(x)=a^2-1\] or is it \[f(x)=x^2-1\]

  12. anonymous
    • one year ago
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    freckles, yes.

  13. anonymous
    • one year ago
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    what it says is what it is. I'm just not understanding how to solve it.

  14. freckles
    • one year ago
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    whatever they are just add f(x) and g(x) to find (f+g)(x)

  15. anonymous
    • one year ago
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    okay

  16. freckles
    • one year ago
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    have you found (f+g)(x)?

  17. anonymous
    • one year ago
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    nope I've been trying but i don't know what I'm doing

  18. freckles
    • one year ago
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    do you know how to add f(x) and g(x) that is do you know how to add a^2-1 and a/(a-1) ?

  19. anonymous
    • one year ago
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    no

  20. freckles
    • one year ago
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    \[(f+g)(x)=f(x)+g(x) \\ =a^2-1+\frac{a}{a-1}\] that is it unless you are asked to combine the fractions

  21. anonymous
    • one year ago
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    no i wasn't.

  22. freckles
    • one year ago
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    try this one: \[(f-g)(x)=f(x)-g(x)\]

  23. anonymous
    • one year ago
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    no clue

  24. freckles
    • one year ago
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    f(x)=a^2-1 and g(x)=a/(a-1) all you have to do is replace f(x) with a^2-1 and g(x) with a/(a-1)

  25. anonymous
    • one year ago
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    what exactly am i doing with this? and should the answer be a equation?

  26. anonymous
    • one year ago
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    i understand none of this and it gets frustrating

  27. freckles
    • one year ago
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    the expression should be an expression of x in this case the expression of x is a constant expression since the expression is really an expression in terms of a

  28. anonymous
    • one year ago
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    I'm attempting to follow

  29. freckles
    • one year ago
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    pretend f(x)=5 and g(x)=6 (f+g)(x)=f(x)+g(x)=5+6=11 you are just replacing f(x) with 5 and g(x) with 6 (f-g)(x)=f(x)-g(x)=5-6=-1 I just replaced f(x) with 5 and g(x) with 6 (f/g)(x)=f(x)/g(x)=5/6 again I just replaced f(x) with 5 and g(x) with 6

  30. freckles
    • one year ago
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    (f*g)(x)=f(x)*g(x)=5*6=30 again I just replaced f(x) with 5 since in my example I let f(x)=5 and I just replaced g(x) with 6 since in my example I let g(x)=6

  31. anonymous
    • one year ago
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    ok

  32. freckles
    • one year ago
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    ok now you try finding (f-g)(x) for your f(x) and g(x)

  33. freckles
    • one year ago
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    remember (f-g)(x) is the same as f(x)-g(x)

  34. freckles
    • one year ago
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    just replace f(x) with a^2-1 and g(x) with a/(a-1)

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