Find the points on the graph of the function that are closest to the given point. f(x) = x^2 − 5, (0, −4)

Question

It asks for a smaller and larger x value.

Answer 1

The distance formula is d = √[(x₂ - x₁)² + (y₂ - y₁)²]. We can define one of the points as (0, -4); let's make that (x₂, y₂) making the distance equation into this: d = √[(0 - x₁)² + (-4 - y₁)²] or d = √[x₁² + (-4 - y₁)²]. The other point (x₁, y₁) is on the graph of f(x) = x² - 5 making the point (x, x² - 5) so we can plug that in to get:

d = √[x² + (-4 - (x² - 5))²] or

d = √[x² + (-4 - x² + 5)²] or

d = √[x² + (-x² + 1)²] or

d = √(x² + x⁴ - 2x² + 1) or

d = √(x⁴ - x² + 1)

To minimize this, we take the derivative and set it equal to zero. To take the derivative, it might be easier to think of it as:

d = (x⁴ - x² + 1)^1/2 then d' (using the chain rule) is:

d' = 1/2((x⁴ - x² + 1)^-1/2(4x³ - 2x) or

d' = (4x³ - 2x)/[2√(x⁴ - x² + 1)]

Setting it equal to zero:

(4x³ - 2x)/[2√(x⁴ - x² + 1)] = 0 can only be true when the numerator equals zero so:

4x³ - 2x = 0

2x(2x² - 1) = 0

x = 0 or 2x² - 1 = 0 making x = ± √(1/2) or ±√2/2

For x = √2/2, y = (√2/2)² - 5 = -9/2 so the point is (√2/2, -9/2) (this is the larger x value)

For x = -√2/2, y = (-√2/2)² - 5 = -9/2 so the point is (-√2/2, -9/2) (this is the smaller x value)

Note: It turns out that the critical point x = 0 is a local maximum.