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midpoint
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inverse of f(x)=pi-arccos(2x+1)vertices y=x^2-2x-24range of f(x)=sqrt(x^2-6x+8)domain of f(x)=ln(7-x)critical f(x)=(ln(x))/(x^7)
Frequently Asked Questions (FAQ)
What is the midpoint (-2,-4),(4,-4) ?
The midpoint (-2,-4),(4,-4) is (1,-4)