Problem 71 - Algebra and Trigonometry

Question

Minimizing a Distance When we seek a minimum or maximum value of a function, it is sometimes easier to work with a simpler function instead. (a) Suppose \(g(x)=\sqrt{f(x)},\) where \(f(x) \geq 0\) for all \(x\) . Explain why the local minima and maxima of \(f\) and \(g\) occur at the same values of \(x .\) (b) Let \(g(x)\) be the distance between the point \((3,0)\) and the point \(\left(x, x^{2}\right)\) on the graph of the parabola \(y=x^{2}\) Express \(g\) as a function of \(x .\) (c) Find the minimum value of the function \(g\) that you found in part (b). Use the principle described in part (a) to simplify your work.

Step-by-Step Solution

Answer

The minimum value of g(x) is \( \sqrt{5} \) at \( x = 1 \).

Step 1: Understanding the Relationship Between f(x) and g(x)

We are given that \( g(x) = \sqrt{f(x)} \), where \( f(x) \geq 0 \) for all \( x \). The derivative of \( g \), \( g'(x) \), is \( \frac{d}{dx} \sqrt{f(x)} = \frac{1}{2\sqrt{f(x)}}f'(x) \). Since \( \sqrt{f(x)} \) is always non-negative, the critical points, where \( g'(x) = 0 \), occur when \( f'(x) = 0 \). Thus, local minima and maxima occur at the same points for both \( f(x) \) and \( g(x) \).

Step 2: Expressing the Distance as a Function of x

The function \( g(x) \) represents the distance between the point \((3, 0)\) and a point \((x, x^2)\) on the parabola \( y = x^2 \). This distance can be expressed as \( g(x) = \sqrt{(x - 3)^2 + (x^2)^2} \), using the distance formula between two points.

Step 3: Simplifying the Function Using f(x)

To find the minimum of \( g(x) \), we can work with \( f(x) = (x - 3)^2 + x^4 \), since minimizing \( \sqrt{f(x)} \) is the same as minimizing \( f(x) \). Our goal is to find the value of \( x \) that minimizes \( f(x) \).

Step 4: Finding the Derivative of f(x)

Calculate the derivative \( f'(x) \). We have: \( f(x) = x^4 + (x-3)^2 = x^4 + x^2 - 6x + 9 \). The derivative is: \( f'(x) = 4x^3 + 2x - 6 \).

Step 5: Solving for Critical Points

Set \( f'(x) = 0 \) to find critical points: \( 4x^3 + 2x - 6 = 0 \). Solving this may involve factoring or using numerical methods. By testing, we find \( x = 1 \) satisfies the equation, providing a critical point.

Step 6: Verifying the Minimum

Check whether \( x = 1 \) is a minimum by evaluating the second derivative \( f''(x) = 12x^2 + 2 \). Since \( f''(1) = 14 \) (which is positive), \( x = 1 \) is a local minimum.

Step 7: Calculating the Minimum Value of g(x)

The minimum value of \( g(x) \) is calculated by substituting \( x = 1 \) into \( g(x) = \sqrt{(x-3)^2 + x^4} \). Therefore, \( g(1) = \sqrt{(1-3)^2 + 1^4} = \sqrt{4 + 1} = \sqrt{5} \).

Key Concepts

Functions

Functions are a fundamental concept in mathematics, describing a relationship between a set of inputs and a set of possible outputs. A function generally assigns exactly one output to each input. In the context of the problem discussed, we are working with functions like \( g(x) \) and \( f(x) \). These functions take a value \( x \) and produce another value by applying specific mathematical formulas directly tied to the points on a graph or a given calculation. Understanding functions involves:

• Identifying the input and output of the function.
• Understanding how changes to the input affect the output.
• Translating words into mathematical expressions, like using the distance formula to express \( g(x) \).

In this exercise, \( g(x) \) was the distance between the point \((3,0)\) and a point on the parabola \(y=x^2\). We expressed \(g(x)\) in terms of \(x\) to solve the problem of minimizing the distance, which meant finding a value of \(x\) that makes \(g(x)\) or equivalently \(\sqrt{f(x)}\) the smallest.

Local Minima and Maxima

Local minima and maxima are points on a graph where the function reaches its lowest or highest value within a small, specific section. These points are crucial in optimization problems, where one seeks to find the smallest or largest values of a function within a given range. This includes minimizing distance problems, a common scenario in calculus. To identify local minima or maxima, a critical step is to find points where the first derivative of the function equals zero, \( g'(x) = 0 \). This indicates potential peaks or valleys. Since the original exercise involved simplifying \( g(x) = \sqrt{f(x)} \), we focused on the critical points of \( f(x) \). Here's how:

• Compute the derivative of \( f(x) \) and set it equal to zero.
• Solve the resulting equation to find critical points, where the slope is zero.
• Use the second derivative to determine if these points are indeed minima (if positive) or maxima (if negative).

In our example, we found \(x = 1\) is a critical point for \(f(x) = (x-3)^2 + x^4\), and using the second derivative test, determined it is a local minimum.

Derivatives

Derivatives offer a way to understand the rate at which a function's output changes with respect to changes in its input. Essentially, derivatives help us analyze the behavior of functions. In calculus, they are instrumental for determining critical points, which provide insight into the local minimum and maximum values of a function. Let's break down some important points about derivatives:

• The first derivative \( f'(x) \) provides information about the slope or the rate of change of the function \( f(x) \).
• Setting the first derivative to zero helps identify critical points, potential candidates for local minima or maxima.
• The second derivative \( f''(x) \) gives us insight into the curvature of the function. A positive value suggests a local minimum, while a negative value indicates a local maximum.

In the given problem, we calculated \( f'(x) = 4x^3 + 2x - 6 \) and set it to zero to find critical points. Then, we used \( f''(x) = 12x^2 + 2 \) to verify the nature of the critical point at \( x = 1 \). This method affirmed that the function had a local minimum at this value of \( x \). By understanding derivatives, we simplified the task of minimizing the distance function \( g(x) \).