Questions: Let P=(x, y) be a point on the graph of y=sqrt(x). (a) Express the distance d from P to the point (4,0) as a function of x. (b) Use a graphing utility to graph d=d(x). (c) For what value(s) of x is d smallest? (d) What is the smallest distance?

Solution Steps

The distance \( d \) between the point \( P = (x, y) \) on the graph of \( y = \sqrt{x} \) and the point \( (4, 0) \) is given by the distance formula: \[ d = \sqrt{(x - 4)^2 + (y - 0)^2} \] Since \( y = \sqrt{x} \), substitute \( y \) into the formula: \[ d = \sqrt{(x - 4)^2 + (\sqrt{x})^2} \] Simplify the expression: \[ d = \sqrt{(x - 4)^2 + x} \]

To graph \( d(x) = \sqrt{(x - 4)^2 + x} \), input the function into a graphing utility. The graph will show the relationship between \( x \) and the distance \( d \).

To find the value(s) of \( x \) that minimize \( d \), take the derivative of \( d(x) \) with respect to \( x \) and set it equal to zero: \[ \frac{d}{dx} \left( \sqrt{(x - 4)^2 + x} \right) = 0 \] First, simplify the expression inside the square root: \[ d(x) = \sqrt{x^2 - 8x + 16 + x} = \sqrt{x^2 - 7x + 16} \] Now, compute the derivative: \[ \frac{d}{dx} \left( \sqrt{x^2 - 7x + 16} \right) = \frac{2x - 7}{2\sqrt{x^2 - 7x + 16}} \] Set the derivative equal to zero: \[ \frac{2x - 7}{2\sqrt{x^2 - 7x + 16}} = 0 \] Solve for \( x \): \[ 2x - 7 = 0 \implies x = \frac{7}{2} \]

Final Answer