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anonymous

  • 5 years ago

Hi, I have problem with a question in increasing and decreasing functions. The question is: Find the intervals of increase or decrease for the function: g(x) = 3x^4 - 16x^3 + 6x^2 + 72x + 8 I've already done the derivative: g(x) = 12x^3 - 48x^2 + 12X + 72 Can anyone help me with the other steps and the answer please?

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  1. anonymous
    • 5 years ago
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    Solve for x, then make intervals from - infinity to x values to positive infinity. Plug in values to see if it is + or -

  2. anonymous
    • 5 years ago
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    Thanks for answering! I've actually tried solving for it..I tried taking the 12 out, didn't work; then tried taking the 12x out, still didn't work..and several other tries... Could you please show me the steps so that I can know what my problems are? :)

  3. anonymous
    • 5 years ago
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    ok try this. first of all find the zeros \[12x^3-48x^2+12x+72=0\] \[x^3-4x^2+x-6=0\]

  4. anonymous
    • 5 years ago
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    typo \[x^3-4x^2+x+6=0\]

  5. anonymous
    • 5 years ago
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    in general it is hard to find the zeros of a cubic polynomial but i guessed it would be easy or you can't do it. x = 1 does not work, but x = -1 does by inspection. so you know -1 is a zero and so it factors as \[(x+1)(something)\]

  6. anonymous
    • 5 years ago
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    find the "something" by division got get \[(x+1)(x^2-5x+6)=0\]

  7. anonymous
    • 5 years ago
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    \[(x+1)(x-3)(x-2)=0\] zeros are -1,2 and 3. now you have the zeros of the cubic polynomial. since the leading coefficient of that cubic is positive, you should have a good idea what it looks like. it crosses the x - axis at -1, 2 and 3 so it must be negative until it gets to -1, then positive between -1 and 2, then negative between 2 and 3 and finally positive from 3 on.

  8. anonymous
    • 5 years ago
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    which means your original function is decreasing on \[(-\infty, -1)\] increasing on \[(-1,2)\] decreasing on \[(2,3)\] and increasing on \[(3,\infty)\] which is precisely what a fourth degree poly with positive leading coefficient should look like.

  9. anonymous
    • 5 years ago
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    Oh my god! Thanks so much satellite73!!! That's the perfect answer! But I still have a question: other than trying numbers, is there any other way that takes less time like simplifying it? I've been trying to simply it like how I do other questions but didn't work on this...

  10. anonymous
    • 5 years ago
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    well you have a cubic polynomial for a derivative. of course it makes it easier to divide by 12. after that it is hard to find the zeros.

  11. anonymous
    • 5 years ago
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    so i just guessed at 1 and then at -1. -1 worked. but in real life the zeroes could be a mess, or two could be complex or who knows. obviously this problems was cooked up so that the zeros would be more or less easy to find. but no, there is not great way to find the zeroes of a cubic polynomial, although there is a formula for it. like the quadratic formula only maybe a page long and very annoying. just guess!

  12. anonymous
    • 5 years ago
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    I see. So how should I write the steps on the exam? Is there an appropriate way to write it out?

  13. anonymous
    • 5 years ago
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    write it like i did. you may want to show how to divide by x+1. you can use synthetic division.

  14. anonymous
    • 5 years ago
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    How did you get (x+1)(x^2−5x+6)? I forgot how to do the division...

  15. anonymous
    • 5 years ago
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    You have to find your old pre-cal notes on long division.

  16. anonymous
    • 5 years ago
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    Oh no:( I don't have them anymore... Would you mind showing me how to do it?

  17. anonymous
    • 5 years ago
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    It is difficult to show here, difficult to write the long division process. Hopefully Satellite or someone has a scanner and can put up an attachment of the work. It is like high school long division, only with letters.

  18. anonymous
    • 5 years ago
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    use synthetic division. easy example here, almost identical to the one we did. http://www.purplemath.com/modules/synthdiv.htm

  19. anonymous
    • 5 years ago
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    Thanks so much! You really helped me out! Now I can go back to study for my exam:)!! Hopefully I'll see you soon here again! (when I need help...but I'll try helping if I can give a hand as well) Thanks again!! :)

  20. anonymous
    • 5 years ago
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    i will be here. if you have a specific question try satellite7.openstudy@gmail.com

  21. anonymous
    • 5 years ago
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    ooops satellite73.openstudy@gmail.com

  22. anonymous
    • 5 years ago
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    thanks:)

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