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MarcLeclair

General question about derivative.. When I take my second derivative I usually end up with crazy numbers and I'm asked for the Inflection point An example of what happens to me is with this function: x^3/(x^2-16) I end up with maybe some x^7 and x^5 at the second derivative and its way too much for me to figure out the inflection point... any help?

  • one year ago
  • one year ago

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  1. bear4343
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    I have to write this in parts so bear with me. d/dx(x^3/(-16+x^2)) = (x^2 (x^2-48))/(x^2-16)^2

    • one year ago
  2. MarcLeclair
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    its okay i got that part its just what comes afterward is insanely long and I can't find the inflection point afterwards

    • one year ago
  3. bear4343
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    (32 x (x^2+48))/(x^2-16)^3 is the second derivative

    • one year ago
  4. bear4343
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    do you get how I got to where I am?

    • one year ago
  5. MarcLeclair
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    Nope, I don't understand how you simplified it, I used a website to get there.. I keep g etting insane numbers

    • one year ago
  6. bear4343
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    I used the quotiant rule. This is the way i learned to remember it. low d high minus high d low and down below the low squareds go.

    • one year ago
  7. bear4343
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    youalsohave to be careful because you also have to do the product rule

    • one year ago
  8. MarcLeclair
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    well i mean after your first step this is what i got : (x^4-48x^2)'(x^2-16)^2-(x^4-48x^2)(x^2-16)^2 / (x^2-16)^4 and then on the numerator to find my inflection point at the end i think i had something like 160x^5- somethingx^3... :/

    • one year ago
  9. bear4343
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    well lets start with an inflection point, the inflection point is where the concavity of the original function changes.

    • one year ago
  10. MarcLeclair
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    yeah where the second derivative is = 0. But I mean I can't do that i have 3 diferent x :/ especially at such high powers

    • one year ago
  11. bear4343
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    what i always do in my mind when solving a problem like this I have to imagine what the graph would look like

    • one year ago
  12. MarcLeclair
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    the problem is that I'm solving the graph at the moment, to me its hard picturing cubic over parabolas, never drew them and I'm new to graph sketching. So when come the time to do the derivative I don't understand how people simpliy that much like you just did

    • one year ago
  13. bear4343
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    I can quickly tell the only point where f" i think also by looking at where you can make it zero because I Know it has to pass through zero when I get the second derivative. If youtake the second derivative and set it equal to zero you can then find the inflection point

    • one year ago
  14. MarcLeclair
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    yeah but with what I get its nearly impossible to get to it because i get this: (x^4-48x^2)'(x^2-16)^2-(x^4-48x^2)(x^2-16)^2' which then turns out as (4x^3-96x)(x^2-16) - 2(x^4-48x^2)(2x) / (x^2-16)^3 and the numerator becomes crazy 4x^5-64x^3-96x^3-1536x - 2(2x^4-96x^3) and I have so many leftovers that I can't deal with

    • one year ago
  15. MarcLeclair
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    oh nevermidn i got it you simplified it, I'm really stupid today x.x thanks a lot bear :)

    • one year ago
  16. bear4343
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    Just copied from above {d/dx(x^3/(-16+x^2)) = (x^2 (x^2-48))/(x^2-16)^2 I simplified it on the right once i found the derivative because i don't want to have to deal with the bigger numbers.

    • one year ago
  17. bear4343
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    Your welcome, if you need any help in the future with calculus let me know.

    • one year ago
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