Problem 47 - University Calculus: Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Critical points are at \( x = 4 \) and \( x = 1 \).

Step 1: Understand Critical Points

Critical points occur where the first derivative of a function is equal to zero or does not exist. We first need to find the derivative of the function.

Step 2: Find the First Derivative

The function is \( f(x) = x(4-x)^3 \). Apply the product rule to differentiate, where \( u = x \) and \( v = (4-x)^3 \). The product rule is \((uv)' = u'v + uv'\).

Step 3: Differentiate Each Part

Calculate \( u' = 1 \) because the derivative of \( x \) is \( 1 \). Next, differentiate \( v = (4-x)^3 \) using the chain rule: \( v' = 3(4-x)^2(-1) = -3(4-x)^2 \).

Step 4: Apply the Product Rule

Substitute into the product rule: \[ f'(x) = 1 \cdot (4-x)^3 + x \cdot (-3(4-x)^2) \]This simplifies to: \[ f'(x) = (4-x)^3 - 3x(4-x)^2 \]

Step 5: Simplify the Derivative

Factor out \((4-x)^2\) from the expression:\[ f'(x) = (4-x)^2[(4-x) - 3x] \]Thus, \[ f'(x) = (4-x)^2(4 - 4x) \] which simplifies to \[ f'(x) = (4-x)^2(4x - 4) \].

Step 6: Set the Derivative to Zero

To find critical points, set \( f'(x) = 0 \):\((4-x)^2(4x - 4) = 0\). This gives us two equations: 1. \( (4-x)^2 = 0 \)2. \( 4x - 4 = 0 \).

Step 7: Solve for Critical Points

The first equation \( (4-x)^2 = 0 \) solves to \( x = 4 \). The second equation \( 4x - 4 = 0 \) solves to \( x = 1 \). Therefore, the critical points are \( x = 4 \) and \( x = 1 \).

Key Concepts

Differentiation

Differentiation is a key concept in calculus, primarily used to determine how a function changes at any given point. It is the process of finding the derivative of a function, which represents the slope of the tangent line to the curve of the function at a particular point. This technique helps in identifying where a function increases, decreases, or reaches a peak or a valley, known as critical points.

To perform differentiation, you apply rules like the product rule and the chain rule, depending on the structure of the function. For instance, if a function is made up of a product of two sub-functions, you would use the product rule to differentiate. Similarly, if a function involves a composition of functions, use the chain rule. Differentiation transforms a function into a form that reveals more about its behavior and characteristics.

Product Rule

The product rule is an essential technique in differentiation used when dealing with functions that are products of two or more simpler functions. The formula for the product rule is:

• ext{If} \, f(x) = u(x)v(x), \text{then the derivative is } f'(x) = u'(x)v(x) + u(x)v'(x).

This rule allows us to differentiate a multiplication of functions with ease.

In the original exercise, the function to differentiate was \( f(x) = x(4-x)^3 \). Here, \( u = x \) and \( v = (4-x)^3 \). First, differentiate \( u \), which is simple since the derivative of \( x \) is \( 1 \). Next, finding the derivative of \( v \) requires the chain rule, which we will discuss next. Once you have both \( u' \) and \( v' \), apply the product rule formula to find the derivative of the original function.

Using the product rule simplifies the process of differentiation, ensuring that each part of the function is accurately considered in the derivative.

Chain Rule

The chain rule is a powerful rule in calculus used to differentiate composite functions, meaning functions inside of other functions. The chain rule formula is:

• ext{If} \, f(x) = g(h(x)),\text{then the derivative is } f'(x) = g'(h(x)) imes h'(x).

It allows us to break down complex functions into simpler parts for easier differentiation.

In the context of the given problem, to differentiate \( v = (4-x)^3 \), we recognize it as a composite function. Let \( g(u) = u^3 \) and \( h(x) = 4-x \). Applying the chain rule involves finding \( g'(u) = 3u^2 \) and \( h'(x) = -1 \). Substitute \( h(x) \) back to get \( g'(4-x) = 3(4-x)^2 \). Finally, multiply with \( h'(x) \) to get \( v' = -3(4-x)^2 \).

This process, empowered by the chain rule, ensures every part of the function is correctly differentiated, leading to an accurate derivative for complex functions.