anonymous
  • anonymous
Calculus help needed! Will medal and fan!!
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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?
Plasmataco
  • Plasmataco
im only 13 but sure, I can try.
Plasmataco
  • Plasmataco
http://media.education2020.com/contentengine/tools/Calculators/graphing-calculator.html

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Plasmataco
  • Plasmataco
type in your equations and get the answer!
ybarrap
  • ybarrap
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
ybarrap
  • ybarrap
$$ x^2=2^x\\ 2\ln x=x\ln2\\ \cfrac{\ln x}{x}=\cfrac{\ln2}{2}\\ \ln x \exp^{-\ln x}=\cfrac{\ln2}{2}\\ -\ln x \exp^{-\ln x}=-\cfrac{\ln2}{2}=W\left(-\cfrac{\ln 2}{2}\right)\\ -\ln x=W\left(-\cfrac{\ln2}{2}\right)\\ x=\exp^{-W\left(-\cfrac{\ln2}{2}\right)}\\ =\frac{-2W\left(\cfrac{-\ln2 }{2}\right)}{\ln 2}=2 $$ 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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.