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anonymous

  • one year ago

If g(x) = sqrt(x-3) and k(x) = x^2 + 5 then determine (kg)(x).

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  1. rhr12
    • one year ago
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    (k-g)(5) means plug 5 into both equations, then divide the answer of k(x) by g(x) (g o k)(7) means plug 7 the equation k(x) and then plug the answer you get from k(x) into g(x)

  2. rhr12
    • one year ago
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    (k - g)(5) =(x²+5)–x–3 = x²-x+2 = 5²-5+2 = 22

  3. anonymous
    • one year ago
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    \[(kg)(x)=(x^2+5)(\sqrt{x ^{-3})}\]

  4. rhr12
    • one year ago
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    hope it helps

  5. anonymous
    • one year ago
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    Where did you get (k-g) @rhr12? It says to multiply k and g. Were you just giving an example?

  6. anonymous
    • one year ago
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    I agree with saseal that \[(kg)(x) = k(x)g(x) = (x^2+5)\sqrt{x-3}\]

  7. anonymous
    • one year ago
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    So what would you plug in for x?

  8. anonymous
    • one year ago
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    Or would you have to multiply that first?

  9. anonymous
    • one year ago
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    \[\sqrt{x-3}=(x-3)^{\frac{ 1 }{ 2 }}\]

  10. anonymous
    • one year ago
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    theres nothing to plug in, theres no value given in the question

  11. anonymous
    • one year ago
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    So I have to multiply it....Couldn't I just square both equations to get rid of the square root?

  12. anonymous
    • one year ago
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    i dont think you can multiply, just leave it that way

  13. anonymous
    • one year ago
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    I don't know think I'd bother multiplying it out, if its actually possible. If I wanted to evaluate (kg) at say x=10, I'd just use it as a normal function and substitute x = 10 everywhere. i.e.\[(kg)(10) = (10^2+5)\sqrt{10-7} = 105\sqrt{7}\]

  14. anonymous
    • one year ago
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    So just write (x^2 + 5)(sqrt(x - 3))?

  15. anonymous
    • one year ago
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    yea

  16. anonymous
    • one year ago
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    It should be \[\sqrt{10-3} \text{ instead of } \sqrt{10-7}\] and yep that is what I'd write

  17. anonymous
    • one year ago
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    Alright thanks!

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