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anonymous

  • one year ago

Calculus help needed! Will medal and fan!!

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  1. anonymous
    • one year ago
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    Where f(x)=-x^2 and g(x)=-2^x, I am looking for the points of intersection. So far, I have started this by setting -2^x=-x^2 to look for x. I worked it to xlog2(2)=2log2(x) and then x(1)=2log2(x) and then x/2=log2(x) and then x/2=ln(x)/ln(2) but now feel as if I am working in a circle... eventually, I reduced it to x=e^(xln(2)/2), however, this still has x on both sides. Help, please?

  2. Plasmataco
    • one year ago
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    im only 13 but sure, I can try.

  3. Plasmataco
    • one year ago
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    http://media.education2020.com/contentengine/tools/Calculators/graphing-calculator.html

  4. Plasmataco
    • one year ago
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    type in your equations and get the answer!

  5. ybarrap
    • one year ago
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    Looks like you will need the Product Log Function - https://en.wikipedia.org/wiki/Lambert_W_function#Example_1 http://www.wolframalpha.com/input/?i=-x%5E2%3D-2%5Ex

  6. ybarrap
    • one year ago
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    $$ x^2=2^x\\ 2\ln x=x\ln2\\ \cfrac{\ln x}{x}=\cfrac{\ln2}{2}\\ \ln x \exp^{-\ln x}=\cfrac{\ln2}{2}\\ -\ln x \exp^{-\ln x}=-\cfrac{\ln2}{2}=W\left(-\cfrac{\ln 2}{2}\right)\\ -\ln x=W\left(-\cfrac{\ln2}{2}\right)\\ x=\exp^{-W\left(-\cfrac{\ln2}{2}\right)}\\ =\frac{-2W\left(\cfrac{-\ln2 }{2}\right)}{\ln 2}=2 $$ Where \(W()\) is the Lambert W function - https://en.wikipedia.org/wiki/Lambert_W_function Another solution is, since x can be positive or negative: $$ =\frac{-2W\left(\cfrac{\ln2 }{2}\right)}{\ln 2}\approx-0.77 $$ Now check if 2^2 is another solution by plugging in to original equation. If so, that will be a third solution.

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