Consider the function. ƒ(x) = 12 - x^2 on the closed interval [0, √12]. The curve y = f(x) is drawn in the figure below (blue). A point (x, y) is on the curve.

Question

Consider the function.

ƒ(x) = 12 - x² on the closed interval [0, √12].

The curve y = f(x) is drawn in the figure below (blue). A point (x, y) is on the curve.

Write a function to compute the distance d from the point (0, 6) to the point (x, y) as a function of x.

d =

What point (x, y) on the curve y = f(x) is closest to the point (0, 6)?

x =

y =

Answer 1

d = √ [(x-0)² + (y - 6)²] = √[x² + (12-x²-6)² ] = √[x² + (6-x²)²] = √(x⁴ -11x² + 36)

d is smallest when S = x⁴ - 11x² + 36 is smallest

Set S' = 0 to get 4x³ - 22x = 0

2x(2x² - 11) = 0

x = 0 or x = √ (11/2)

The minimum occurs at a critical point or at an endpoint

If x = 0, then y = 12. and d = 6

If x = √(11/2), then y = 12 - x² = 12 - 11/2 = 6.5 and d = √(121/4 -121/2 + 36) < 6

If x = √12, then y = 0 and d = √(144 - 132 + 36) > 6

So x = √(11/2) and y = 6,5