Problem 47 - Thomas Calculus in SI Units

Question

Determine all critical points for each function. $$y=x^{2}-32 \sqrt{x}$$

Step-by-Step Solution

Answer

The function has one critical point at \( x = 4 \).

Step 1: Find the Derivative

To find the critical points, we first need to find the derivative of the function. The function is given by \( y = x^2 - 32\sqrt{x} \). Rewrite \( \sqrt{x} \) as \( x^{1/2} \). So, the function becomes \( y = x^2 - 32x^{1/2} \). The derivative of \( x^2 \) is \( 2x \). The derivative of \( 32x^{1/2} \) is \( 32 \times \frac{1}{2}x^{-1/2} = 16x^{-1/2} \). This gives us the derivative: \[ y' = 2x - \frac{16}{\sqrt{x}} \]

Step 2: Set Derivative to Zero

Critical points occur when the derivative is equal to zero, so we set the derivative to zero and solve for \( x \): \[ 2x - \frac{16}{\sqrt{x}} = 0 \]

Step 3: Solve the Equation

Multiply through by \( \sqrt{x} \) to clear the fraction: \[ 2x\sqrt{x} = 16 \]Let \( u = \sqrt{x} \), then \( x = u^2 \) and \( 2u^3 = 16 \). Divide both sides by 2: \[ u^3 = 8 \] Taking the cube root of both sides, we get \( u = 2 \). Since \( u = \sqrt{x} \), substituting back, we get \( \sqrt{x} = 2 \), hence \( x = 4 \).

Step 4: Verify Derivative Equals Zero at Solution

Check the derivative at \( x = 4 \). Substitute \( x = 4 \) into the derivative: \[ y' = 2(4) - \frac{16}{\sqrt{4}} = 8 - 8 = 0 \]Since the derivative equals zero, \( x = 4 \) is indeed a critical point.

Key Concepts

Derivative

A derivative measures how a function changes when the input changes. Specifically, it's the rate of change or slope of the function at any given point. To find a derivative, we often use rules of differentiation, such as the power rule. In the exercise, the derivative of the function \( y = x^2 - 32\sqrt{x} \) is calculated by first rewriting \( \sqrt{x} \) as \( x^{1/2} \). This simplification helps in applying the differentiation rules, making it easier to compute derivatives using known formulas.
The steps include applying the power rule: for an expression \( x^n \), the derivative is \( nx^{n-1} \). Therefore, the derivative of \( x^2 \) becomes \( 2x \) and of \( 32x^{1/2} \) as \( 16x^{-1/2} \). Thus, the resulting derivative \( y' = 2x - \frac{16}{\sqrt{x}} \) provides insights into where the slope of the original function is zero, indicating potential critical points for further analysis.

Function

Functions in mathematics represent relationships between sets of inputs and outputs. They are fundamental in calculus and can be visualized as graphs. Each input in the domain of a function maps to exactly one output in its range. The given exercise concerns a specific function: \( y = x^2 - 32\sqrt{x} \).
Functions are important because they help us model real-world situations. Analyzing their behavior—like finding critical points—can reveal where a function reaches a local maximum or minimum. These points are found by setting the derivative equal to zero and solving for the variable. Understanding these concepts in the context of functions is essential for comprehending broader mathematical theories and applications.

Calculus

Calculus is a branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. It provides tools for analyzing trends and behaviors of functions as they approach different values. The exercise highlights one of calculus' core functions: finding critical points to explore where a function's rate of change is zero.
Critical points, such as those found in the exercise, signify where a function changes direction. This can reveal local maximums, minimums, or points of inflection. By understanding calculus, students gain insights into dynamic systems, trends, and predictable patterns that help solve complex problems in fields ranging from engineering to economics.

Mathematics Education

Mathematics education involves teaching and learning strategies to ensure that learners of all levels can understand and apply mathematical concepts effectively. By focusing on foundational topics like derivatives and functions, students build essential skills for advanced studies in calculus and other mathematical areas.
Understanding how to derive functions, find critical points, and analyze graphs is crucial for students. It's not just about computing answers but developing reasoning skills, critical thinking, and problem-solving abilities. Strengthening these skills in mathematics classes can have a lasting impact, helping students navigate diverse academic and real-world challenges with confidence.