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midpoint
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symmetry 1/(x^2)domain of f(x)= r/((r-3)^2)*(r^2-9)/(2r)critical f(x)=(x-2)e^xinverse of f(x)=-(x-3)^2+1line (1,1),(4,-1)
Frequently Asked Questions (FAQ)
What is the midpoint (1,5),(-3,-1) ?
The midpoint (1,5),(-3,-1) is (-1,2)