Question

Find the point on the graph of the function that is closest to the given point. $$ \frac{\text { Function }}{f(x)=\sqrt{x}} \quad \frac{\text { Point }}{(4,0)} $$

Short answer

Therefore, the point on the graph of the function that is nearest to the point (4,0) is \((7/2, \sqrt{7/2})\).

Step-by-step solution

Formulate the distance equation

The distance between any two points (x1,y1) and (x2,y2) can be defined using the distance formula, this is true for any values in a Cartesian plane. The formula is: \(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\). For our problem, this results in the function: \(D(x) = \sqrt{(x-4)^2+(\sqrt{x}-0)^2}\). Expanding the square root would make differentiation easier later.

Simplify the function

Simplify the function to make further steps easier. After expanding the square root, the equation becomes: \(D(x) = \sqrt{x^2 - 8x + 16 + x}\).

Differentiate the D(x) function

We'll find where D(x) is at a minimum by finding the point where its derivative is 0. However, differentiating the square root can be messy, so it's simpler to minimize \(D^2(x)\), which will give us the same result. So let's differentiate \(D^2(x) = x^2 - 8x + 16 + x\). The derivative is \(D'(x) = 2x - 8 + 1\). This derivative function gives us the rate of change of the distance for each value of x.

Find the critical points

Setting the first derivative equal to zero will give us the critical points. Solve the equation \(2x - 8 + 1 = 0\). Solving this equation will give a critical point at \(x = 7/2\) or 3.5.

Analyze the critical points

At \(x = 7/2\), evaluate the graph to confirm if the distance is at minimum. We only have one critical point, so no need to compare with others. Substituting \(x = 7/2\) into \(f(x) = \sqrt{x}\) yields \(f(7/2) = \sqrt{7/2}\). So the point on the graph of the given function closest to (4,0) is \((7/2, \sqrt{7/2})\).

Key concepts

Distance Formula

Understanding the distance formula is crucial when tackling calculus optimization problems that involve finding the shortest distance between two points. It's a staple in geometry, and often emerges in optimization exercises. The formula is given as
\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]
where \((x_1, y_1)\) and \((x_2, y_2)\) are the coordinates of the two points. The formula itself is derived from the Pythagorean theorem, reflecting the distance as the hypotenuse of a right triangle formed by the two points in the Cartesian plane. In the context of optimization, it plays a vital role by transforming a spatial relationship into a function that can be analyzed with calculus.

Function Differentiation

Function differentiation is an indispensable tool in calculus, particularly in optimization problems. It’s the process of finding the derivative of a function, which gives the rate at which the function's value changes as its input changes. When we differentiate a function, we're essentially looking for a new function, called the derivative, which provides the slopes of the tangents to the curves of the original function at any point.
For example, given a function \(f(x)\), its derivative is denoted as \(f'(x)\) or \(\frac{df}{dx}\). In optimization problems, finding the derivative of a distance function, like \(D(x)\) in the given example, allows us to determine where the function's value increases or decreases, and ultimately, where it reaches either a maximum or minimum – this is the heart of the optimization process.

Critical Points Analysis

A critical point occurs where the first derivative of a function is either zero or undefined. These points are potential candidates for local maxima or minima of the function. Once found, critical points must be analyzed to determine the nature of these potential extremes. This analysis usually involves taking the second derivative of the function or examining the first derivative's behavior around the critical points.
In optimization problems such as the one described, we set the first derivative equal to zero and solve for the variable to find the critical points. Subsequently, we substitute these points back into the original function to see what values they yield. For example, the critical point found at \( x = 7/2 \) is then used to find the minimum distance to the point on the graph. By examining critical points, we address the central question of optimization: What are the best possible values under the given conditions?