Given that \(f(5)=5\) and it crosses \(y=0\) at the following points:
\(x=200, x=100,x=0,x=-100,x=-200\)
Since the crossing points of this function is rational, I deduct that the function is factorizable.
Perform reverse-factorization:
The function need to be in this form:
\(g(x)=(x+200)(x+100)(x+0)(x-100)(x-200)\)
For it to have roots at \(x=200, x=100, x=0, x=-100,x=-200\)
There are a total of five zero-points at:
\(1.x=-200 , y=0\)
\(2.x=-100 , y=0\)
\(3.x=0 , y=0\space(Origin)\)
\(4.x=100 , y=0\)
\(5.x=200, y=0\)
There are a total of four critical points at:
Local Maxima:
\(1.x=\sqrt{15000-1000\sqrt{145}} , y=200000000\left(\sqrt{5}+5\sqrt{29}\right)\sqrt{30-2\sqrt{145}}\)
\(2.x=-\sqrt{15000+1000\sqrt{145}} , y=-200000000\left(\sqrt{5}-5\sqrt{29}\right)\sqrt{30+2\sqrt{145}}\)
Local Minima:
\(3.x=\sqrt{15000+1000\sqrt{145}} , y=\left(\sqrt{5}-5\sqrt{29}\right)\sqrt{30+2\sqrt{145}}\)
\(4.x=-\sqrt{15000-1000\sqrt{145}} , y=-200000000\left(\sqrt{5}+5\sqrt{29}\right)\sqrt{30-2\sqrt{145}}\)
Since \(f(5)=5\), Just divide every y-value of maximas and minimas above by a factor of \(g(5)/5=398750625\)
\(f(x)=\frac{1}{398750625}\left(x+200\right)\left(x+100\right)x\left(x-100\right)\left(x-200\right)\)
Q.E.D.
(I bet you just randomly typed the numbers in, didn't you? (Because the \(x\) and \(y\) values are pretty ugly to be honest))
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