anonymous
  • anonymous
Any line of the form y = mx will intersect the curve y = x/ (x^2 + 10) in precisely three points provided 0 < m < B for some number B. What is the value of B? When the line y = mx intersects the curve in three points, this line and the curve y = x/ (x^2+10) will enclose a region consisting of two parts. Find the total area of the enclosed region. Your answer should be an expression involving m.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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IrishBoy123
  • IrishBoy123
for \(mx = \dfrac{x}{x^2 + 10}\), ignoring trivial solution leaves \(x = \dfrac{1}{x^2 + 10}\) which you can solve using the quadratic formula, looking for real solutions for x.
anonymous
  • anonymous
Alright, I got the first part down. How do I begin the second part? When the line y = mx intersects the curve in three points, this line and the curve y = x/ (x^2+10) will enclose a region consisting of two parts. Find the total area of the enclosed region. Your answer should be an expression involving m. I know that I need to solve for x, where the intersections are. I have one point where x = 0, but I do not know how to find the other two x's to do the integral.
IrishBoy123
  • IrishBoy123
you got \(x = \pm \sqrt{\frac{1}{m} - 10}\) for the non zero roots?!?! i have plotted it and its symmetric, even.....so use symmetry and why not try \(\large 2 \int\limits_{0}^{\sqrt{\frac{1}{m} - 10}} \quad dx \quad \left( \dfrac{x}{x^2 + 10} - mx \right)\) twice this?! http://www.wolframalpha.com/input/?i=int_%7B0%7D%5E%7Bsqrt%281%2Fm+-+10%29%7D+++x%2F%28x%5E2+%2B+10%29+-+m*x+++dx&dataset=&equal=Submit

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