anonymous
  • anonymous
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?
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
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
anonymous
  • anonymous
its okay i got that part its just what comes afterward is insanely long and I can't find the inflection point afterwards
anonymous
  • anonymous
(32 x (x^2+48))/(x^2-16)^3 is the second derivative

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anonymous
  • anonymous
do you get how I got to where I am?
anonymous
  • anonymous
Nope, I don't understand how you simplified it, I used a website to get there.. I keep g etting insane numbers
anonymous
  • anonymous
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.
anonymous
  • anonymous
youalsohave to be careful because you also have to do the product rule
anonymous
  • anonymous
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... :/
anonymous
  • anonymous
well lets start with an inflection point, the inflection point is where the concavity of the original function changes.
anonymous
  • anonymous
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
anonymous
  • anonymous
what i always do in my mind when solving a problem like this I have to imagine what the graph would look like
anonymous
  • anonymous
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
anonymous
  • anonymous
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
anonymous
  • anonymous
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
anonymous
  • anonymous
oh nevermidn i got it you simplified it, I'm really stupid today x.x thanks a lot bear :)
anonymous
  • anonymous
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.
anonymous
  • anonymous
Your welcome, if you need any help in the future with calculus let me know.

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