Problem 43 - Thomas Calculus in SI Units

Question

Determine all critical points for each function. $$f(x)=x(4-x)^{3}$$

Step-by-Step Solution

Answer

The critical points are at \( x = 2 \) and \( x = 4 \).

Step 1: Identify the Function to Differentiate

The given function is \( f(x) = x(4-x)^3 \). We need to find the derivative of this function to locate the critical points.

Step 2: Apply the Product Rule

The function \( f(x) = x(4-x)^3 \) is a product of two functions: \( u(x) = x \) and \( v(x) = (4-x)^3 \). The derivative using the product rule is \( f'(x) = u'(x)v(x) + u(x)v'(x) \).

Step 3: Differentiate Each Component

First, find the derivatives:\( u'(x) = 1 \) because the derivative of \( x \) is 1.To find \( v'(x) = (4-x)^3 \), use the chain rule. Let \( g(x) = 4-x \), then \( v(x) = g(x)^3 \). The derivative \( g'(x) = -1 \) and using the chain rule \( v'(x) = 3(4-x)^2(-1) = -3(4-x)^2 \).

Step 4: Substitute Derivatives into Product Rule

Using the product rule from Step 2, we have:\( f'(x) = 1 \cdot (4-x)^3 + x \cdot (-3(4-x)^2) \).Simplifying, we get:\( f'(x) = (4-x)^3 - 3x(4-x)^2 \).

Step 5: Factor the Derivative

Factor the common term \((4-x)^2\) from \( f'(x) \):\( f'(x) = (4-x)^2((4-x) - 3x) \).Simplify further to get:\( f'(x) = (4-x)^2(-2x + 4) \).

Step 6: Solve for the Critical Points

Set \( f'(x) = 0 \):\((4-x)^2(-2x + 4) = 0\).This equation is satisfied if either \((4-x)^2 = 0\) or \(-2x + 4 = 0\).Solving \((4-x)^2 = 0\), we find \( x = 4 \).Solving \(-2x + 4 = 0\), we find \( x = 2 \).

Step 7: Conclude the Critical Points

The critical points occur at \( x = 2 \) and \( x = 4 \). These are the x-values where the first derivative is zero, indicating potential local maxima or minima.

Key Concepts

Derivatives

Derivatives are a fundamental concept in calculus. They represent the rate at which a function changes at any given point. This can be thought of as the slope of the tangent line to the graph of the function. In the exercise, we begin by determining the derivative of the function \( f(x) = x(4-x)^3 \). The derivative, \( f'(x) \), helps us identify critical points where the function might have local maxima or minima. When finding the derivative of complex functions, we often involve techniques like the product rule and chain rule. These help in breaking down the function into manageable parts for differentiation. In general, understanding derivatives is key to analyzing how functions behave.

Product Rule

The product rule is a method used in calculus for differentiating products of two functions. If you have two functions, \( u(x) \) and \( v(x) \), the product rule states that the derivative of their product is given by:

• \( (uv)' = u'v + uv' \)

This rule is essential when you encounter a function like \( f(x) = x(4-x)^3 \) from the exercise. Here, \( u(x) = x \) and \( v(x) = (4-x)^3 \). By applying the product rule, you find the derivative \( f'(x) \) by differentiating each component separately and then combining them according to the rule. This step is crucial for finding critical points, where the derivative equals zero.

Chain Rule

The chain rule is used when differentiating composite functions. A composite function is a function inside another function, like \( v(x) = (4-x)^3 \). In these cases, the chain rule provides a systematic way to find the derivative of the entire function.It works as follows:

• Identify the outer function and the inner function.
• Differentiate the outer function with respect to the inner function.
• Multiply that result by the derivative of the inner function.

In the exercise, you apply the chain rule to find \( v'(x) \). First, you set \( g(x) = 4-x \) as the inner function, then differentiate \( (g(x))^3 \), and finally multiply by \( g'(x) \). This gives \( v'(x) = 3(4-x)^2(-1) \), which helps solve for critical points. Understanding the chain rule allows you to deal with more complex differentiation scenarios.

Local Maxima and Minima

Once you've found the derivative, the next step is to identify the critical points, which can signal potential local maxima or minima. Critical points occur where the derivative \( f'(x) \) equals zero or where it does not exist.In the exercise, solving \( (4-x)^2(-2x + 4) = 0 \) allows you to find the values of \( x \) that are critical points—in this case, \( x = 2 \) and \( x = 4 \). At these points, the function has interesting behavior. A local maximum is where the function reaches a peak, while a local minimum is where it hits a low point. To confirm whether a critical point is a maximum or minimum, you would typically examine the second derivative or analyze the sign changes of the first derivative around the critical points.