Calculus
posted by Michelle .
Find the point on the graph of y=2x4 that is closest to the point (1,3). (Optimization equation)

D^2=f(x)= (x1)^2+((2x4)3)^2
f'(x)= 2(x1)(1)+2(2x7)(2)
0= 2x2+(4x14)2
0= 2x2+8x28
0= 10x30
30=10x
x=3
y=2(3)4
y=2
(3,2)
the square over D doesn't matter.
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