SOLUTION: solve algebraically e^x = -x

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Question 404418: solve algebraically e^x = -x
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
It's quite clear that no x+%3E=+0 can satisfy the equation. (e%5Ex+%3E+0, but -x+%3C=+0). Considering the graphs of both e%5Ex and -x, there is only one solution (or intersection point), and it lies between -1 and -1/2.
(This is because e%5E%28-1%29+-+1+%3C+0 and e%5E%28-1%2F2%29+-+1%2F2+%3E+0.
Now e%5Ex++=+1+%2B+x+%2B+x%5E2%2F2 for small values of x.
So to get an approximate value for the root, solve
-x+=+1%2Bx+%2B+x%5E2%2F2
==> x%5E2%2F2+%2B+2x+%2B+1++=0, or
x%5E2+%2B+4x+%2B+2+=+0.
==> x+=+%28-4+%2B-+sqrt%28+16-8+%29%29%2F2+
x+=+%28-4+%2B-+2sqrt%282%29%29%2F2+
x+=+-2+%2B-+sqrt%282%29%29+. Eliminating the bottom value the answer is
x+=+-2+%2B+sqrt%282%29%29+