Problem 38 - Calculus Early Transcendentals

Question

Use any method to find the relative extrema of the function \(f .\) $$ f(x)=x(x-4)^{3} $$

Step-by-Step Solution

Answer

A local minimum occurs at \(x = 1\).

Step 1: Find the First Derivative

To find the relative extrema, we first need to find the first derivative of the function. The function is given by \[ f(x) = x(x-4)^3. \]Use the product rule, which states that \((uv)' = u'v + uv'\), with \(u = x\) and \(v = (x-4)^3\). Thus, \[ f'(x) = (1)(x-4)^3 + x(3(x-4)^2). \]Simplifying this, we express the derivative as \[ f'(x) = (x-4)^3 + 3x(x-4)^2. \]

Step 2: Simplify the First Derivative

Factor out the common term \((x-4)^2\) from the expression for the first derivative:\[ f'(x) = (x-4)^2(x-4 + 3x). \]This simplifies to:\[ f'(x) = (x-4)^2(4x-4). \]

Step 3: Set the First Derivative Equal to Zero

To find critical points, set the first derivative equal to zero:\[ (x-4)^2(4x-4) = 0. \]This equation is satisfied if:1. \((x-4)^2 = 0 \Rightarrow x = 4,\) or 2. \(4x-4 = 0 \Rightarrow 4x = 4 \Rightarrow x = 1.\)

Step 4: Determine the Second Derivative

To determine the nature of the critical points, find the second derivative. Using the quotient rule and simplification, the expression for the second derivative is\[ f''(x) = 2(x-4)(4x-4) + 4(x-4)^2. \]Simplify this to evaluate at the critical points.

Step 5: Evaluate the Second Derivative at Critical Points

Calculate the second derivative at the critical points:1. For \(x = 4\), use the expression from Step 4 to find \(f''(4).\) The calculations will show \[ f''(4) = 12\cdot(4-4) + 0 = 0. \] This requires further investigation as it may denote an inflection point.2. For \(x = 1\), find \(f''(1).\) Calculations result in \[ f''(1) = 16 - 12 = 4. \] Since \(f''(1) > 0,\) a local minimum occurs at \(x = 1.\)

Key Concepts

Understanding the First Derivative

To find the relative extrema of a function, one of the first steps involves finding its first derivative. The first derivative, denoted as \( f'(x) \), helps us understand the rate at which the function \( f(x) \) changes at a given point. For the given function \( f(x) = x(x-4)^3 \), we utilize the product rule, which helps when dealing with the product of two functions. The product rule states that if we have two functions \( u(x) \) and \( v(x) \), their derivative is \( (uv)' = u'v + uv' \). This rule simplifies the differentiation process

• For \( u = x \), the derivative \( u' \) is 1.
• For \( v = (x-4)^3 \), using the chain rule, \( v' \) is \( 3(x-4)^2 \).

Applying the product rule, we can find \( f'(x) \) as \( (x-4)^3 + 3x(x-4)^2 \). Simplifying gives \( f'(x) = (x-4)^2(4x-4) \). This expression is crucial since setting \( f'(x) = 0 \) helps us find the critical points, where the slope of the function is zero, indicating potential relative extrema. It's essential to approach derivatives with a structured methodology, often breaking them into smaller parts, and applying rules like the product or chain rule effectively.

Exploring the Second Derivative

Once we have determined the critical points using the first derivative, we use the second derivative for more details. The second derivative, denoted \( f''(x) \), provides information about the curvature of the function. It tells us how the rate of change itself is changing. Calculating the second derivative involves differentiating \( f'(x) \) again. For \( f(x) = x(x-4)^3 \), using the expression \( f'(x) \) derived earlier, we find \( f''(x) = 2(x-4)(4x-4) + 4(x-4)^2 \). Evaluating this at the critical points reveals more about the nature of those points:

• If \( f''(x) > 0 \), the function is concave up, indicating a local minimum.
• If \( f''(x) < 0 \), the function is concave down, indicating a local maximum.
• If \( f''(x) = 0 \), the test is inconclusive, and further analysis may be necessary.

For instance, at \( x = 1 \), \( f''(1) = 4 \) demonstrates a local minimum, as the second derivative is positive. Whereas at \( x = 4 \), \( f''(4) = 0 \) does not provide clear information by itself, suggesting it could be an inflection point.The second derivative test is a powerful tool to confirm the nature of critical points and ensure accuracy in identifying extremas.

Determining Critical Points

Critical points are where the function experiences changes in direction, potentially indicating maxima, minima, or points of inflection. To find them, the first derivative \( f'(x) \) is set to zero as these correspond to places where the slope of the tangent to the function is zero. For the function \( f(x) = x(x-4)^3 \), the derivative \( f'(x) = (x-4)^2(4x-4) \) is set to zero. Solving \( (x-4)^2(4x-4) = 0 \) yields \( x = 4 \) and \( x = 1 \) as solutions. This illustrates that critical points can be found through straightforward algebraic manipulation. These points are particularly significant:

• Analyzing \( f'(x) \) presents the conditions for discovering critical points as places where derivatives vanish or become undefined.
• Investigating the surroundings of each point using derivatives confirms whether they are indeed extremes.

Critical points mark locations of potential significance in a function’s graph, guiding our investigation into the nature of these points with the first and second derivative tests. Connecting the derivatives' behavior allows deeper insights into the function's graph and supports identifying relative extrema.