Calculus
posted by Mark .
Find the points on the curve y=x2+2 closest to the point (0,3). Enter the coordinate with the smallest xvalue first and round to the nearest 4 decimal places.

Calculus 
Steve
y = x^2+2
y' = 2x
So, at (h,k) y'(h) = 2h
So, you want a line with slope 1/(2h) passing through (0,3)
y3 = 1/(2h) x
h^2+2  3 = 1/2
h^2 = 1/2
h = 1/√2
so, (1/√2,5/2) is closest to (0,3)
1/2 +
Or, consider the distance formula. We have
z^2 = (x0)^2 + (y3)^2
= x^2 + (x^2+23)^2
= x^2 + x^42x^2+1
= x^4  x^2 + 1
2z z' = 4x^3  2x
z' = x(2x^21)/z
z'=0 at x=0, x = ±1/√2
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