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ashking1

  • one year ago

Find (f + g)(x).

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  1. ashking1
    • one year ago
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    \[ g= \sqrt{6x-9} \]

  2. ashking1
    • one year ago
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    f=\[\sqrt{6x+9}\]

  3. anonymous
    • one year ago
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    \[(f+g)(x)=f(x)+g(x)\] Use this formula, everything else is given

  4. anonymous
    • one year ago
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    what subject is this for?

  5. ashking1
    • one year ago
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    precal i think i got it is the answer 6x

  6. anonymous
    • one year ago
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    I believe so.

  7. ashking1
    • one year ago
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    can you help me with one more

  8. anonymous
    • one year ago
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    sure

  9. ashking1
    • one year ago
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    Determine algebraically whether the function is even, odd, or neither even nor odd. f(x)= x+ 4/x

  10. anonymous
    • one year ago
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    what are the answer options just want to check

  11. ashking1
    • one year ago
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    even odd or neither

  12. anonymous
    • one year ago
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    f(x)= x+4/x in my opinion would be odd mainly because its the opposite of the 4/x

  13. anonymous
    • one year ago
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    excuse me if im wrong i havent been studying this

  14. freckles
    • one year ago
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    (f+g)(x)=f(x)+g(x) don't know how you got 6x if really is that \[f(x)=\sqrt{6x+9} \text{ and } g(x)=\sqrt{6x-9}\]

  15. freckles
    • one year ago
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    just replace f(x) with sqrt(6x+9) and replace g(x) with sqrt(6x-9)

  16. freckles
    • one year ago
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    also to determine if a function is odd or even (or neither odd or even) the first step is to plug in -x

  17. freckles
    • one year ago
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    if you receive f(-x)=f(x), then f is even if you receive f(-x)=-f(x), then f is odd

  18. ashking1
    • one year ago
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    so it would be odd am i correct

  19. freckles
    • one year ago
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    \[f(x)=x+\frac{4}{x} \\ \text{ plug \in } -x \\ f(-x)=-x+\frac{4}{-x} \\ f(-x)=-(x+\frac{4}{x}) \\ f(-x)=-f(x) \\ \text{ yep } f \text{ is odd } \\ \text{ unless you meant } f(x)=\frac{x+4}{x} \\ \text{ then the story is a bit different }\]

  20. freckles
    • one year ago
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    you would still plug in -x of course

  21. anonymous
    • one year ago
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    yes

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