Find the point on the graph of f(x) = square root of x that is closet to the point (4,0).?

Question

Find the point on the graph of f(x) = square root of x that is closet to the point (4,0).

a. (7/2 , square root of 14/2)

b. (0,0)

c. (4,2)

d. (3/2 , square root of 6/2)

Answer 1

Distance between (4,0) and (x,√x) is a minimum. So is distance squared

= (x₁-x₂)² + (y₁-y₂)²

= (4-x)² + (0-√x)²

= x²-8x+16+x

= x²-7x+16

This is an up opening parabola whose minimum x = -b/2a = 7/2

y = f(x) = √x = √(7/2)

(7/2,√(7/2)) = (7/2,√14/2) = (a)

Answer 2

(3.5,√3.5)

 

Answer 3

Let's call that:

y = √x

There probably is a better way to do it, but let's try it this way. Equation showing the distance between two points:

d = √[(x₁ - x₂)² + (y₁ - y₂)²]

Let's call x₁ and y₁ the unknowns x and y, and use the given point for x₂ and y₂. So now we have:

d = √[(x - 4)² + (y - 0)²]

d = √(x² - 8x + 16 + y²)

We know that y = √x, so y² = x. Substitute that and simplify:

d = √(x² - 8x + 16 + x)

d = √(x² - 7x + 16)

If we get the first derivative of that and solve for the zero, we'll get the point where d is at a minimum, which is what you are looking for:

dd/dx = (2x - 7) / √(x² - 7x + 16)

Solve for the zero:

0 = (2x - 7) / √(x² - 7x + 16)

Multiply both sides by the denominator:

0 = 2x - 7

And this is now easy:

7 = 2x

7/2 = x

So we have the x, now we need to find the y:

y = √x

y = √(7/2)

rationalize denominator:

y = √(14) / 2

So this is the point that is closest to the given point:

(7/2, √(14)/2) (answer A)