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Let $$f(x)={x^3-14x^2+7x+203\over(x-3)(8-x)}$$ I want to find the two solutions of $$f(x)f''(x)=(f'(x))^2,\qquad3\le x\le8$$

This is the first time i am using maple, and i cannot get the graph to work out.

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    Which functions "intersect"? Which graph are you wanting to work out?2012-04-26
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    Reading your post is a nightmare. Could you please write mathematics in some less "computer science" fashion?2012-04-26
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    edited, i removed the first and second Der. to make the post cleaner. I believe you all should be able to calculate these, so its not important for me question.2012-04-26
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    @Siminore better?2012-04-26
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    What are the functions you are trying to find the intersection of?????2012-04-26
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    Are you looking at $$F(x)=\frac{x^3-14x^2+7x+203}{(x-3)(8-x)}$$ and its derivatives $F'(x)$ and $F''(x)$ and finding the points at which $F(x)$, $F'(x)$, and $F''(x)$ intersect?2012-04-26
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    $F1 = |F(x)*F''(x)|$ and $F2 = F'(x)^2$2012-04-26
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    Ah. OK. So we have: $F(x)$, $F_1(x)=|F(x)F''(x)|$, and $F_2(x)=F'(x)^2$? I am asking just to make sure I fully understand.2012-04-26
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    Yes! and i want to know where F1 intersects F2. In fact, i want to know where (over 3, 8) is F1 < F22012-04-26
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    Does this help? I have to leave and cannot continue the problem: \begin{align} F(x)&=-\frac{x^3-14x^2+7x+203}{(x-8)(x-3)}\\ F'(x)&=-\frac{(x^2-11x+49)^2}{(x-8)^2(x-3)^2}\\ F''(x)&=\frac{50(2x^3-33x^2+219x-539)}{(x-8)^3(x-3)^3}\\ |F(x)F''(x)|&=\frac{50 (2 x-11) (x^2-11x+49) (x^3-14x^2+7x+203)}{(x-8)^4(x-3)^4}\\ F'(x)^2&=\frac{(x^2-11x+49)^4}{(x-8)^4(x-3)^4}\\ \end{align}2012-04-26
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    In the light of the discussions here, I have taken the liberty of cleaning up the statement of the question, and re-tagging.2012-04-27

2 Answers 2

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Using Maple 16 ...

computation

Note $F1 > -10$ and $F2 > 50$, so there is no need to consider the absolute value of $F1$ for intersections.

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The first derivative is $3 x^2-28 x+ \frac {203}{5 (x-8)^2}-\frac {203}{5 (x-3)^2}+7$ and the second is $\frac {14}5 \left(\frac{29}{(x-3)^3}-\frac{29}{(x-8)^3}-10\right )+6 x$ (both from Alpha). Unfortunately, the entry box isn't large enough for your whole problem, but a lot of the denominators will cancel.