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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Sign up to join this communityLet $$f(x)={x^3-14x^2+7x+203\over(x-3)(8-x)}$$ 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.
Using Maple 16 ...
Note $F1 > -10$ and $F2 > 50$, so there is no need to consider the absolute value of $F1$ for intersections.
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.