Problem 12 - Calculus Early Transcendentals

Question

Locate the critical points and identify which critical points are stationary points. $$ f(x)=x^{2}(x-1)^{2 / 3} $$

Step-by-Step Solution

Answer

Critical points are found at \( x = 0 \), determining stationary nature by examining algebraic solutions remaining after division simplification.

Step 1: Find the First Derivative

To locate the critical points, we first need to find the first derivative of the function. The function given is \( f(x) = x^2 (x-1)^{2/3} \). We use the product rule and chain rule to differentiate:1. \( u(x) = x^2 \) and \( v(x) = (x-1)^{2/3} \).2. \( u'(x) = 2x \).3. \( v'(x) = \frac{2}{3}(x-1)^{-1/3} \).Using the product rule, \( f'(x) = u'(x)v(x) + u(x)v'(x) \), we get:\[ f'(x) = 2x(x-1)^{2/3} + x^2 \left( \frac{2}{3}(x-1)^{-1/3} \right) \].

Step 2: Simplify the First Derivative

Combine terms to simplify \( f'(x) \):1. The first term is \( 2x(x-1)^{2/3} = 2x(x-1)^{2/3} \).2. The second term is \( \frac{2}{3}x^2(x-1)^{-1/3} = \frac{2x^2}{3(x-1)^{1/3}} \).To have a common denominator, multiply the first term by \( \frac{(x-1)^{1/3}}{(x-1)^{1/3}} \):\[ f'(x) = \frac{2x(x-1) + \frac{2}{3}x^2}{(x-1)^{1/3}} = \frac{2x(x-1)^{1/3}(x-1) + \frac{2}{3}x^2}{(x-1)^{1/3}} \].

Step 3: Set the Derivative to Zero

Find where \( f'(x) = 0 \) to locate critical points:\( 2x(x-1)^{1/3}(x-1) + \frac{2}{3}x^2 = 0 \). Factor out common terms and solve:\[ \frac{2x(x(x-1)+\frac{x}{3})}{(x-1)^{1/3}} = 0 \].Since \( x = 0 \) or \( \frac{x(x-1) + x/3}{(x-1)^{1/3}} = 0 \), solving yields the terms when \( x = 0 \) and solve the equation \( x(x-1) + \frac{x}{3} = 0 \). This simplifies to potential solutions for \( x \).

Step 4: Identify Stationary Points

Analyze where the critical points are stationary. A stationary point occurs wherever \( f'(x) = 0 \) and doesn't leave terms undefined, such as division by zero.Solutions from the previous step like \( x = 0 \) and other algebraic solutions such as points from dividing polynomial terms identify places where changes cease, marking stationary points.

Key Concepts

Derivative

The derivative of a function is one of the core concepts in calculus.It represents the rate at which a function is changing at any given point.For the function given, \( f(x) = x^2 (x-1)^{2/3} \), finding the derivative involves using some specific rules of differentiation such as the product and chain rules.

To locate critical points, we need the first derivative.Critical points occur where this derivative equals zero or is undefined.In calculus, these points help us understand where a function is stationary or could change direction.

Finding the derivative accurately is crucial for solving problems involving rate of change and critical points.In this problem, differentiating \( f(x) \) gives us key information about where potential changes in the function's graph occur.

Product Rule

When faced with differentiating a function that is a product of two functions, the product rule comes in handy.This rule states that the derivative of a product \( u(x) \cdot v(x) \) is given by \( u'(x)v(x) + u(x)v'(x) \).

In our original exercise, \( f(x) = x^2 (x-1)^{2/3} \), we used the product rule because \( f(x) \) is the product of \( u(x) = x^2 \) and \( v(x) = (x-1)^{2/3} \).Here, \( u'(x) = 2x \) and \( v(x) \) needs to be differentiated using another technique called the chain rule.

Applying the product rule helps us break down complex problems into manageable steps, making the differentiation process easier to follow and execute.For complex functions, understanding and applying the product rule is an invaluable skill.

Chain Rule

The chain rule is another key tool in calculus, used to differentiate composite functions.A composite function is one where one function is inside another, such as \( v(x) = (x-1)^{2/3} \).The chain rule states that the derivative of a composite function \( g(f(x)) \) is \( g'(f(x)) \cdot f'(x) \).

In our task, when differentiating \( (x-1)^{2/3} \), we treat \( (x-1) \) as the inner function and \( y^{2/3} \) as the outer function.First, differentiate the outer function with respect to the inner function, which gives \( \frac{2}{3}(x-1)^{-1/3} \).Then, multiply this by the derivative of the inner function, \((x-1)' = 1\).

The chain rule simplifies the process of working with nested functions, allowing for an accurate calculation of their rates of change.

Stationary Points

Stationary points are locations on a graph where the derivative is zero, indicating no change in the function's value locally.They are critical to understanding the behavior of a function as they might correspond to local maxima, minima, or points of inflection.

From our original problem, after setting \( f'(x) \) to zero, we calculated possible values for \( x \) where the function is stationary.By solving the equation \( f'(x) = 0 \), we determine points like \( x = 0 \) and other possible terms that indicate the function's rate of change ceases at those points.

Stationary points provide valuable insights into the graph of a function, highlighting regions of potential turning points and areas where the function might not increase or decrease.