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haleyelizabeth2017

  • one year ago

For which interval(s) is f(x)=x^3+x^2-5x-6 increasing or decreasing?

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  1. anonymous
    • one year ago
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    increasing maybe

  2. haleyelizabeth2017
    • one year ago
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    It's not just an "increasing/decreasing" answer lol I have to find the intervals

  3. Owlcoffee
    • one year ago
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    There is something interesting we can actually prove, let's generalize for some point \(a(x_a,y_a)\) on the plane, we have \(f'(a)>0\). This must mean that by definition: \(\lim_{x \rightarrow a} \frac{ f(x)-f(x_a) }{ x-x_a }\). Then by the theorem of conservation of sign we know that there must exist a \(\delta >0\) that belongs to the interval \(\left| x-a \right|<\delta\) such that \[\frac{ f(x)-f(x_a) }{ x-x_a }>0\] From here we can conclude two things: (1) if \(x<x_a\) then \(x-x_a<0\) which implies that \(f(x)-f(x_a)<0\) then \(f(x)<f(x_a)\) (2) if \(x>x_a\) then \(x-x_a>0\) which implies that \(f(x)-f(x_a)>0\) then \(f(x)>f(x_a)\) by the very definition of increasing or decreasing function we can conclude that function "f" is increasing on the point \(a(x_a,y_a)\). \(if: f'(a)>0 \rightarrow \) f increases on "a". This process is completely analogical to f'(a)<0. Returning to the excercise: \[f(x)=x^3+x^2-5x-6\] What we will do is use the theorem i proved to you earlier, and we will need the derivative of the function in order to apply it: \[f'(x)=3x^2+2x-5\] now, we want to know what makes the derivative zero: \[3x^2+2x-5=0\]

  4. haleyelizabeth2017
    • one year ago
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    -5/3 or -1?

  5. freckles
    • one year ago
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    @Owlcoffee I don't think @haleyelizabeth2017 knows any calculus

  6. haleyelizabeth2017
    • one year ago
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    Lol nope...This would be precalc lol

  7. freckles
    • one year ago
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    Calculus is awesome because you can do these questions without a calculator. And it isn't that bad so I guess next year or semester you will get there.

  8. Owlcoffee
    • one year ago
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    Great, now you have to study the sign of that interval. That happens to the function if x takes values greater -1? And what happens if x takes values lesser than -5/3? The study of sign looks like this: |dw:1443820022212:dw| This means that for the interval \((x; + \infty)\) the function is increasing. for the interval \((-\frac{ 5 }{ 3 };-1)\) the function is decreasing. for the interval \((- \infty ; -\frac{ 5 }{ 3 })\) the function is increasing. you can verify with your calculator.

  9. Owlcoffee
    • one year ago
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    if she's in precalc she must know about derivation, and that's enough to find the variation of a function on the plane @freckles

  10. freckles
    • one year ago
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    I helped her with the previous one. She didn't know derivatives if that is what is meant by the word derivation.

  11. freckles
    • one year ago
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    So that is why I think these are calculator questions instead.

  12. Owlcoffee
    • one year ago
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    I'm sorry, I am from the old school and I did not have those graphing calculators.

  13. freckles
    • one year ago
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    I do like old school stuff. I think calculus is fantastic for these questions and they should probably be saved for calculus class.

  14. Owlcoffee
    • one year ago
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    I personally believe that teachers shouldn't teach with graphing calculators, and start teaching using their brains.

  15. freckles
    • one year ago
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    I totally agree....

  16. Owlcoffee
    • one year ago
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    but yeah, I already stated the answer... So... Let's move on!

  17. freckles
    • one year ago
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    I wasn't trying to put down your way... I was just saying the op is unfamiliar with that way.

  18. freckles
    • one year ago
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    Your way actually is the exact way I started on her previous question.

  19. Owlcoffee
    • one year ago
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    Don't worry about it, I thought she had the knowledge of derivative. So that's why I started like that.

  20. haleyelizabeth2017
    • one year ago
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    Sadly not :(

  21. Owlcoffee
    • one year ago
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    How can someone define increasing or decreasing interval without derivative?... What's wrong with theachers these days?

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