Problem 43 - Thomas Calculus

Question

In Exercises \(41-50,\) determine all critical points for each function. $$ f(x)=x(4-x)^{3} $$

Step-by-Step Solution

Answer

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

Step 1: Find the Derivative

To determine the critical points of the function \(f(x) = x(4-x)^3\), we first need to find its derivative \(f'(x)\). We will use the product rule, which states that \((uv)' = u'v + uv'\), where \(u = x\) and \(v = (4-x)^3\). Calculate the derivative:1. Compute \(u' = 1\).2. Compute \(v'\) using the chain rule: \(v' = 3(4-x)^2 \cdot (-1) = -3(4-x)^2\).Therefore, the derivative is:\[ f'(x) = 1 \cdot (4-x)^3 + x \cdot (-3)(4-x)^2 \]\[ f'(x) = (4-x)^3 - 3x(4-x)^2 \]

Step 2: Simplify the Derivative

Simplify the expression for \(f'(x)\):\[ f'(x) = (4-x)^3 - 3x(4-x)^2 \]Factor out \((4-x)^2\) from both terms:\[ f'(x) = (4-x)^2 ((4-x) - 3x) \]This simplifies to:\[ f'(x) = (4-x)^2 (4-4x) \]

Step 3: Set the Derivative to Zero

To find the critical points, we set \(f'(x) = 0\):\[ (4-x)^2 (4-4x) = 0 \]The solution occurs when either:1. \((4-x)^2 = 0\)2. \(4-4x = 0\)

Step 4: Solve for \(x\)

Solve the equations obtained in Step 3:1. \((4-x)^2 = 0\) implies that \(4-x = 0\). Thus, \(x = 4\).2. \(4-4x = 0\) implies that \(4 = 4x\). Thus, \(x = 1\).The critical points are \(x = 4\) and \(x = 1\).

Step 5: Verify Critical Points

To ensure these are critical points, check if the function is defined at \(x = 1\) and \(x = 4\). The original function \(f(x) = x(4-x)^3\) is a polynomial, hence it is defined for all \(x\) in the real numbers.Thus, both \(x = 1\) and \(x = 4\) are valid critical points for the function.

Key Concepts

Derivative

A derivative represents the rate at which a function changes. Basically, it tells you how much the output of a function changes when you change the input by a tiny amount. This concept is crucial in finding critical points, which are where the function's slope is zero or undefined. To find a derivative, especially for a function that is a product of two terms, the product rule and sometimes the chain rule are handy tools. In the context of our original problem, the derivative of the function given helps us locate these critical points, as they are found where the derivative equals zero.
Understanding derivatives is important because they help you analyze and sketch the graph of a function. With a derivative, you can locate maxima, minima, and points of inflection that tell you key information about the behavior of a function.

Product Rule

The product rule is a very useful technique in calculus, specifically when dealing with functions that are the product of two or more factors. The rule states that the derivative of the product of two functions, say \( u \) and \( v \), is \( (uv)' = u'v + uv' \).
For example, in our given function \( f(x) = x(4-x)^3 \), the choice of \( u = x \) and \( v = (4-x)^3 \) allows us to efficiently compute the derivative.

• First, differentiate \( u \), giving \( u' = 1 \).
• Then differentiate \( v \), requiring the chain rule for handling \( (4-x)^3 \).
• Apply the product rule: multiply \( u' \) by \( v \) and add it to \( u \) times \( v' \).

The product rule is indispensable when you encounter products of functions, as it breaks down a complex problem into simpler parts.

Chain Rule

The chain rule is a fundamental tool in calculus used to differentiate composite functions. It helps when your function is nested within another function. For a function \( v = (4-x)^3 \), the outside function is the cube and the inside function is \( (4-x) \).
Using the chain rule, you take the derivative of the outside function and multiply it by the derivative of the inside function. Here's how it applies:

• The outside function \( (x)^3 \) gives \( 3(x)^2 \).
• The inside function \( (4-x) \) yields \( -1 \) when differentiated.

So, applying the chain rule, we find the derivative \( v' = 3(4-x)^2(-1) = -3(4-x)^2 \).
This approach is key to unraveling complex functions by addressing them in stages, which can make the task of differentiation more manageable.

Polynomial Function

A polynomial function is an expression comprising variables raised to whole number powers and their coefficients. The beauty of polynomial functions is their simplicity and flexibility, as they can be added, subtracted, and multiplied to form new polynomials.
In the context of our exercise, the function \( f(x) = x(4-x)^3 \) is a polynomial, specifically a cubic polynomial once expanded.

• Critical points for polynomials are found by setting the derivative to zero.
• Polynomials are differentiable over their entire domain, which is the real numbers.

For this particular function, the derivative simplified as \( (4-x)^2(4-4x) \) gives us the critical points by solving \( (4-x)^2 = 0 \) and \( 4-4x = 0 \), resulting in critical points at \( x = 4 \) and \( x = 1 \). Polynomial functions are foundational because they model a wide array of real-world problems with flexibility and relative ease.