Question

Find the point on the graph of the function that is closest to the given point. $$ \begin{array}{ll} \text { Function } & \text { Point } \\ f(x)=x^{2} & \left(2, \frac{1}{2}\right) \end{array} $$

Short answer

The point on the graph of the function \(f(x) = x^{2}\) that is closest to the point (2, 1/2) is approximately (1.587, 2.518).

Step-by-step solution

Formulate the Distance Equation

We can use the formula of Euclidean distance to find the distance between the point (2, 1/2) and any point (x, y) on the curve \(y = x^{2}\). The distance formula is: \[D = \sqrt{(x - x_1)^2 + (y - y_1)^2}\]. In this case, the distance between the function, \(y = x^{2}\), and the given point, (2, 1/2) is: \[D = \sqrt{(x - 2)^2 + (x^{2} - 1/2)^2}\]

Differentiate the Equation

To minimize the distance, differentiate the distance equation with respect to x and set it equal to zero. We square the distance to eliminate the square root, simplifying the differentiation. So, differentiate \(D^{2} = (x - 2)^2 + (x^{2} - 1/2)^2\) which gives us: \[2D \cdot D' = 2(x - 2) + 2(x^{2} - 1/2) \cdot 2x\] Cancel out common factor of 2. Then, set the derivative equal to zero: \[D' = (x - 2) + 2x(x^{2} - 1/2) = 0\]

Find the x-coordinate

Solve the equation for x. We get the equation to be: \[x^{3} - x + 2 = 0 \]. This equation doesn't factor nicely, so it must be solved using a numerical approximation method such as the Newton-Raphson method or by using a calculator. We get the answer approximately x = 1.587.

Find the y-coordinate

Substitute x = 1.587 into the original function \(y = x^{2}\) to get the y-coordinate of the point on the curve. Substitution gives us \(y = (1.587)^{2} = 2.518169\). So the coordinates of the point on the curve that is closest to (2, 1/2) is approximately (1.587, 2.518).