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find the x value of the point of the curve y=3x+2 closest to the point (1,0)
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That equation is not a curved line, that equation is a straight line (y=mx+b),
ok...but this is what it says on my test
What class?
calculus I
distance=d=sqrt((1-x)^2+(0-(3x+2)^2) minimizing distance will give you same minimum if you minimize d^2 d^2=(1-x)^2+(3x+2)^2 (d^2)'=..... you finish this i have to go
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you could find the min value between the point (1,0) and (x,3x+2) using the distance formula and then set the derivative of the distance fomula to 0
oh ok. Thnks guys
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