Posted by Anonymous on Sunday, May 25, 2008 at 1:00pm.
There is more than one solution to this problem. Try an polynomial equation of with a derivative that must satsify the two zero points:
f'(x) = (x+3)(x-1) = x^2 +2x -3
f(x) = x^3/3 + x^2 -3x + (any constant)
Note that there is an arbitrary constant, so there are already an infinite number of functions. I could just as well have picked a cosine function f(x) = cos (a*pi*x) with zero slope
f'(x) = - a pi sin a*pi*x = 0
at integer values of x, with a being any integer
(ax = n*pi)
We can't draw graphs for you. You can do that.
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