Closest point on a curve What point on the parabola y = 1 - x² is closest to the point (1, 1)?

Verified step by step guidance

1. To find the point on the parabola y = 1 - x^2 that is closest to the point (1, 1), we need to minimize the distance between a point (x, y) on the parabola and the point (1, 1). The distance D between these two points is given by the formula D = \sqrt{(x - 1)^2 + (y - 1)^2}.
2. Since y = 1 - x^2 for points on the parabola, substitute y in the distance formula to get D = \sqrt{(x - 1)^2 + ((1 - x^2) - 1)^2}. Simplify the expression inside the square root to get D = \sqrt{(x - 1)^2 + (-x^2)^2}.
3. To simplify the problem, minimize the square of the distance instead of the distance itself, as the square root function is monotonically increasing. Define the function f(x) = (x - 1)^2 + (-x^2)^2 and find its minimum value.
4. Differentiate f(x) with respect to x to find the critical points. The derivative f'(x) = 2(x - 1) + 2(-2x)(-x^2) = 2(x - 1) + 4x^3. Set f'(x) = 0 to find the critical points.
5. Solve the equation 2(x - 1) + 4x^3 = 0 for x. This will give you the x-coordinate of the point on the parabola that is closest to (1, 1). Substitute this x value back into y = 1 - x^2 to find the corresponding y-coordinate.