The point on the parabola y = x^2 closest to (0, 1/2) has x-coordinate

0 Squared distance D = x^2 + (x^2 - 1/2)^2. D' = 2x + 2(x^2-1/2)(2x) = 2x[1 + 2(x^2-1/2)] = 2x[2x^2]. Zero at x=0. (At x=0, D = 1/4; smaller than alternatives.)

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Acemy

Optimization

AP Calculus AB· difficulty 4/5

y = x^2

A
$1/2$
B
$0$
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C
$\pm 1$
D
$\pm \sqrt{2}/2$

Explanation

D = x^2 + (x^2 - 1/2)^2

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