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

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)

3 Answers

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  • ?
    Lv 7
    3 years ago

    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)

  • DWRead
    Lv 7
    3 years ago

    (3.5,√3.5)

    Attachment image
  • 3 years ago

    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)

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