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mickey Group Title

f(x)=1/(4+9x^2) - sketch the region bounded by f, the x-axis, and the line x=1 and x=2

  • 3 years ago
  • 3 years ago

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  1. julie Group Title
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    hmmm, well, I think basically this is a second derivative function. Paul has some pretty good notes here on how to use the second derivative to glean more about the shape of a function http://tutorial.math.lamar.edu/Classes/CalcI/ShapeofGraphPtII.aspx

    • 3 years ago
  2. julie Group Title
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    Right now I'm watching this video: http://www.brightstorm.com/math/calculus/applications-of-the-derivative/curve-sketching-with-derivatives

    • 3 years ago
  3. julie Group Title
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    seems to explain it well also. but yeah you need information from both the first and second derivatives in order to correctly sketch the region

    • 3 years ago
  4. julie Group Title
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    been years since I've done so :p

    • 3 years ago
  5. julie Group Title
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    $$f(x)=1/(4+9x ^{2})$$

    • 3 years ago
  6. julie Group Title
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    ^^ function using the equation editor thingy

    • 3 years ago
  7. mickey Group Title
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    Thanks Julie I think that helps me some!

    • 3 years ago
  8. julie Group Title
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    hmmm, honestly going to have to look up the chain rule again lol

    • 3 years ago
  9. julie Group Title
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    to even get the first derivative

    • 3 years ago
  10. julie Group Title
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    hmmm. does anyone remember the chain rule? (looking)

    • 3 years ago
  11. shadowfiend Group Title
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    So we have: $$\frac{1}{4 + 9x^2}$$ Which makes: $$(4 + 9x^2)^{-1}$$ We can apply the chain rule to this by first setting: $$ u = (4 + 9x^2) $$ And: $$\frac{d}{du} u^{-1} = -u^{-2}$$

    • 3 years ago
  12. shadowfiend Group Title
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    But the chain rule says that taking the derivative of u when u consists of a function of x means we have to multiply by the derivative of the inner function: $$ \frac{d}{dx} \left(4 + 9x^2\right) = \frac{d}{dx}4 + \frac{d}{dx}9x^2 $$ We can apply basic power rule and such and get: $$ \frac{d}{dx} \left(4 + 9x^2\right) = 0 + 18x = 18x $$ And our end result is: $$ -\left(4 + 9x^2\right)^-2 \cdot 18x = -\frac{18x}{\left(4 + 9x^2\right)^2} $$

    • 3 years ago
  13. shadowfiend Group Title
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    That's the first derivative, you can apply the same reasoning to get the second derivative, and then use julie's advice to find the shape of the function.

    • 3 years ago
  14. shadowfiend Group Title
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    (Feel free to correct me anyone -- I'm doing this in my head and haven't gotten a chance to double-check myself.)

    • 3 years ago
  15. MD Group Title
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    One way of remembering the Chain Rule in this type of case is to remember that it will be the derivative of the outside function with the inside function left alone times the derivative of the inside function.

    • 3 years ago
  16. MD Group Title
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    So, in this case once you write the fucntion as, $$ (4+9x^{2})^{-1}$$ the "outside function" is the outermost function, i.e. the exponent of "-1" while the "inside function" is then $$4+9x^{2}$$, i.e. the stuff on the inside.

    • 3 years ago
  17. MD Group Title
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    We know that the derivative of say $u^{-1}$ is $-u^{-2}$ so differentiating the outside, while leaving the inside alone would give, $$-\left( 4+9x^{2}\right)^{-2}$$ Then multiply all this by the derivative of the inside, $18x$. Or, putting it all together you get what shadofiend got.

    • 3 years ago
  18. julie Group Title
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    wow thanks MD! very helpful. chain rule is a bit... confusing to decipher without explanation

    • 3 years ago
  19. MD Group Title
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    Yes, it's can be difficult when trying to learn it (or scape the rust off skills long unused :) ). If you recall the outside/inside function way of using the chain rule it can help with a lot of the "simpler" problems. Although I suspect "simpler" is in the eye of the beholder... :)

    • 3 years ago
  20. mickey Group Title
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    okay my text is working again

    • 3 years ago
  21. julie Group Title
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    @MD indeed! @mickey indeed!

    • 3 years ago
  22. shadowfiend Group Title
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    @MD Thanks -- by the way, for inline LaTeX, I'm afraid you have to use \ ( and \ ) (without the spaces) so that we don't start messing up when people start using $ for money :) So \(u^{-1}\) is \(u^{-2}\) should work.

    • 3 years ago
  23. MD Group Title
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    Ah, I was wondering about that. It's been ages since I've seriously used LaTeX and I was thinking I'd misremembered how to do it (got a lot of rust there :) ). I'd forgotten about the other way of doing inline LaTeX. Thanks for reminding me! On a side note a preview button might be good when typing into the box directly, i.e without using the Equation editor....

    • 3 years ago
  24. shadowfiend Group Title
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    Yep. Ideally we'd like the equation editor itself to allow inline equations.

    • 3 years ago
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