Problem 147 - Calculus Volume 1

Question

Find the critical points of the following generic functions. Are they maxima, minima, or neither? State the necessary conditions. \(y=(x-1)^{a}\), given that \(a>1\)

Step-by-Step Solution

Answer

The critical point is at \( x = 1 \), which is likely a point of inflection.

Step 1: Find the first derivative

To locate the critical points of the function, we first need to find the first derivative of \( y = (x-1)^a \). Use the power rule, which states that the derivative of \( x^n \) is \( nx^{n-1} \), and the chain rule. The first derivative is given by: \[ \frac{dy}{dx} = a(x-1)^{a-1} \]

Step 2: Set the first derivative to zero

Critical points occur where the first derivative is zero or undefined. Set \( \frac{dy}{dx} = 0 \): \[ a(x-1)^{a-1} = 0 \] Since \( a > 1 \), \( a eq 0 \). Thus, \[ (x-1)^{a-1} = 0 \]\.

Step 3: Solve for x

To solve \( (x-1)^{a-1} = 0 \) for \( x \), set the expression inside the power equal to zero: \[ x-1 = 0 \] Hence, \[ x = 1 \] This is the critical point.

Step 4: Determine if the critical point is a maximum, minimum, or neither

Calculate the second derivative to apply the second derivative test. The second derivative is given by differentiating the first derivative:\[ \frac{d^2y}{dx^2} = a(a-1)(x-1)^{a-2} \]Evaluate the second derivative at the critical point \( x = 1 \):\[ \frac{d^2y}{dx^2}\bigg|_{x=1} = a(a-1)(1-1)^{a-2} = 0 \]The second derivative test is inconclusive since the result is zero. Further investigation, such as analyzing the sign change of the first derivative or higher derivatives, may be needed, but in general it suggests a point of inflection.

Key Concepts

Derivative

The derivative is a fundamental concept in calculus that represents the rate at which a function's value changes as its input changes. For a given function, the derivative provides insight into its

• slope,
• tangents,
• and behavior at different points.

In essence, the derivative gives us a way to understand how a function behaves locally. When we aim to find the critical points of a function, the first step is to determine its first derivative. This helps us understand where the function may attain a maximum or minimum value. The process involves using rules of differentiation, such as the power rule and chain rule, which we will discuss further.

Power Rule

The power rule is one of the most commonly used techniques in differentiation. It states that for a function of the form \( x^n \), the derivative is determined by \( nx^{n-1} \). This simple rule makes calculating derivatives straightforward for polynomial-like expressions.

In the original exercise, where the function is \( y = (x-1)^a \), the power rule was applied to find the first derivative.

• By treating \((x-1)\) as a single entity, we can raise it to the power of \(a\).
• The derivative then results in \( a(x-1)^{a-1} \).

This expression is crucial because setting it to zero helps identify potential critical points, which are examined in the larger context of the problem.

Chain Rule

The chain rule is essential when differentiating composite functions, meaning functions within functions. It explains how to take derivatives of such complex functions by breaking them into simpler parts.

Suppose you have a function \( y = f(g(x)) \). Then, according to the chain rule, \(rac{dy}{dx} = f'(g(x)) imes g'(x)\).

In the exercise, the function \( y = (x-1)^a \) involves a composite structure because \( x-1 \) is inside the power. To solve it:

• First apply the power rule to the outer function, which is \( u^a \) where \( u = x-1 \).
• Then multiply the result by the derivative of the inside function \( u = x-1 \).

This strategy ensures that the derivative is computed correctly, incorporating changes from both the outer and inner functions.

Second Derivative Test

The second derivative test is employed to determine the nature of a critical point once the first derivative is calculated. By taking the second derivative of a function, one can infer whether the critical point corresponds to a maximum, minimum, or a point of inflection:

• If the second derivative at a critical point is positive, the function has a local minimum there.
• If it is negative, the function has a local maximum.
• If the second derivative is zero, the test is inconclusive, and the point might be an inflection point.

In our exercise, the function's second derivative \( a(a-1)(x-1)^{a-2} \) was evaluated at the critical point\( x = 1 \) and resulted in zero. This indicates the test is inconclusive. Therefore, additional methods, like testing sign changes in the first derivative or exploring higher-order derivatives, might be needed for deeper analysis.