Question

Find the points on the graph of the function that are closest to the given point. \(f(x)=x^{2}, \quad\left(2, \frac{1}{2}\right)\)

Short answer

The points on the graph of the function that are closest to the point \((2, \frac{1}{2})\) are given by \((x_1, f(x_1))\) and \((x_2, f(x_2))\), where \(x_1\) and \(x_2\) are the solutions of the quadratic equation from step 4.

Step-by-step solution

Formulate the Distance Function

Since we're going to find the points on the graph that are closest to a specific point, the first thing we should do is to formulate a distance function. In Euclidean geometry, the distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by \(\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\). Now, replace \(y_1\) with the function \(f(x)\), and \(x_2\), \(y_2\) with the point coordinates. This leads us to the function \(D(x) = \sqrt{(x - 2)^2 + (x^2 - \frac{1}{2})^2}\).

Simplify the Function

Finding derivative of a square root function could be quite complex. In order to avoid this, we recognize that minimizing the square of the distance is equivalent to minimizing the distance itself. Hence, redefine the function as \(D_s(x) = (x - 2)^2 + (x^2 - \frac{1}{2})^2\). This won't affect the point where minimum occurs.

Derive the Function

Differentiate \(D_s(x)\): \(D_s'(x) = 2(x - 2) + 2(x)(2x - \frac{1}{2})\).

Set the Derivative to Zero

To find the points where minimum occurs, set the \(D_s'(x)\) to zero and solve for \(x\). Thus, \(2(x - 2) + 2(x)(2x - \frac{1}{2}) = 0\). This is a quadratic equation, which can be solved by factoring or using the quadratic formula.

Solve for x

The solutions to the quadratic equation from the previous step are \(x_1\) and \(x_2\). Plug these values into \(f(x)\) to get the y-coordinates. Thus, the closest points are \((x_1, f(x_1))\) and \((x_2, f(x_2))\) on the graph.