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anonymous

  • one year ago

Let f(x) = x^2 + x + 2 and g(x) = 2x^2 + 5. Find f(g(x)). Show each step of your work.

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  1. anonymous
    • one year ago
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    @Vocaloid I just need to learn how to do the problem, not the answer.

  2. Nnesha
    • one year ago
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    replace x in f(x) function by g(x)

  3. anonymous
    • one year ago
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    I did that, I got: (2x^2+5)^2+2x^2+2

  4. Nnesha
    • one year ago
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    you forgot something \[ \huge\rm (2x^2+5)^2+\color{blue}{x}+2\] x = 2x^2 + 5

  5. Nnesha
    • one year ago
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    you forgot 5

  6. anonymous
    • one year ago
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    did I? Sorry.

  7. anonymous
    • one year ago
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    So to solve, is it 2x^4 or 4x^4

  8. Nnesha
    • one year ago
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    \[ \huge\rm (2x^2+5)^2+\color{blue}{2x^2 +5}+2\]

  9. Nnesha
    • one year ago
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    \[ \huge\rm \color{reD}{x^2}+\color{blue}{x}+2\] \[ \huge\rm \color{ReD}{(2x^2+5)^2}+\color{blue}{2x^2 +5}+2\] x is 2x^2 + 5

  10. anonymous
    • one year ago
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    Yes, so now we have to square x.

  11. Nnesha
    • one year ago
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    yes( 2x^2 +5)^2 is same as (2x^2 +5) (2x^2 +5) so foil it

  12. anonymous
    • one year ago
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    \[4x^4+20x^2+25\]

  13. anonymous
    • one year ago
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    right?

  14. Nnesha
    • one year ago
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    yes that's right

  15. anonymous
    • one year ago
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    So now we combine like terms?

  16. Nnesha
    • one year ago
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    yesp

  17. anonymous
    • one year ago
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    \[4x^4+20x^2+25+2x^2+5+2\]

  18. anonymous
    • one year ago
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    turns into

  19. anonymous
    • one year ago
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    \[4x^4+22x^2+32\]

  20. anonymous
    • one year ago
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    Right?

  21. Nnesha
    • one year ago
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    yep

  22. anonymous
    • one year ago
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    So now I need to factor?

  23. Nnesha
    • one year ago
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    nope that's ur answer

  24. anonymous
    • one year ago
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    Oh, is it? Good, my problem was I tried factoring that and had no idea how to do it.

  25. anonymous
    • one year ago
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    Thanks! :D

  26. Nnesha
    • one year ago
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    my pleasure and yes that's it they wants us to find f(g(x)) not factoring

  27. anonymous
    • one year ago
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    I fanned, so if I need anything, I'll make sure to tag you.

  28. Nnesha
    • one year ago
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    thanks i'll try my best :=)

  29. anonymous
    • one year ago
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    @Nnesha I have one more problem that seems pretty easy.

  30. anonymous
    • one year ago
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    I think I may know how to solve it but graphing and radicals are my weakpoint.

  31. anonymous
    • one year ago
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    Describe how to transform \[(\sqrt[6]{x ^{5}})^{7}\] into an expression with a rational exponent

  32. Nnesha
    • one year ago
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    alright remember this exponent rule \[\huge\rm \sqrt[m]{x^n} = x^\frac{ n }{ m }\] you can convert root to exponent form

  33. anonymous
    • one year ago
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    Yes I know this, it would be \[(x^\frac{ 5 }{ 6 })^7\]

  34. anonymous
    • one year ago
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    then you multiply 7 and 5?

  35. Nnesha
    • one year ago
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    yep! right you already know

  36. Nnesha
    • one year ago
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    \[\huge\rm (x^m)^n = x^{ m \times n}\]

  37. anonymous
    • one year ago
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    \[x^\frac{ 35 }{ 6 }\]

  38. anonymous
    • one year ago
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    thats a 35/6 btw

  39. Nnesha
    • one year ago
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    yep

  40. anonymous
    • one year ago
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    And that is my answer?

  41. Nnesha
    • one year ago
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    yep right

  42. anonymous
    • one year ago
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    K, thanks. Just wanted to check my answer.

  43. Nnesha
    • one year ago
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    your answer is correct you're a mathematician :=)

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