Problem 19 - Thomas Calculus in SI Units

Question

Identify the coordinates of any local and absolute extreme points and inflection points. Graph the function. $$y=4 x^{3}-x^{4}=x^{3}(4-x)$$

Step-by-Step Solution

Answer

Derivative is \( y' = 4x^2(3-x) \).

Step 1: Find the First Derivative

To find the local extreme points, we need to first find the derivative of the function. The function is given by \[ y = 4x^3 - x^4 = x^3(4-x) \] To find the first derivative, use the product rule: \[ y' = u'v + uv' \]where \( u = x^3 \) and \( v = 4-x \).Differentiating, we get:\[ u' = 3x^2 \]\[ v' = -1 \]Substitute these into the product rule:\[ y' = (3x^2)(4-x) + (x^3)(-1) \]Simplify:\[ y' = 12x^2 - 3x^3 - x^3 \]\[ y' = 12x^2 - 4x^3 \]\[ y' = 4x^2(3-x) \].

Step 2: Find the first derivative

Compute \(f'(x)\) using the appropriate differentiation rules.

Step 3: Find critical points

Set \(f'(x) = 0\) and solve for \(x\). Also check where \(f'(x)\) is undefined.

Step 4: Apply the second derivative test

Compute \(f''(x)\) and evaluate it at each critical point to classify them as local maxima, minima, or neither.

Step 5: Determine the answer

Based on the analysis, identify the required extrema and state the final answer.

Step 6: Conclude with the answer

Derivative is \( y' = 4x^2(3-x) \).

Key Concepts

First Derivative

The first derivative of a function plays a crucial role in calculus, especially in identifying critical points. It provides the rate at which the original function changes as the independent variable, typically designated as \( x \), changes. For the function given in the exercise, \( y = 4x^3 - x^4 = x^3(4-x) \), we need to find its first derivative to investigate its behavior.
First, we apply the product rule, which is necessary when dealing with products of functions. This rule states that the derivative of a product \( u(x)v(x) \) is \( u'v + uv' \). In our exercise, \( u = x^3 \) and \( v = 4-x \). By differentiating these:

• \( u' = 3x^2 \)
• \( v' = -1 \)

Utilizing the product rule, the derivative becomes \( y' = (3x^2)(4-x) + (x^3)(-1) \). Simplifying this expression, we get \( y' = 12x^2 - 4x^3 \). Now, \( y' = 4x^2(3-x) \) fully captures the derivative's form and tells us where the slope of the tangent to the function is zero or undefined, leading to critical analysis of the function's local extremes.

Product Rule

The product rule is a fundamental differentiation tool used when you have a function composed of two multiplied functions, say \( u(x) \) and \( v(x) \). The derivative of this product, according to the product rule, is \( u'v + uv' \). It simplifies the differentiation process for more complex functions.
In our scenario with \( y = x^3(4-x) \), identifying \( u \) and \( v \) as \( x^3 \) and \( 4-x \) respectively, and their derivatives as \( u' = 3x^2 \) and \( v' = -1 \), allows us to apply the product rule efficiently. The calculation becomes straightforward: plugging into the formula, we arrive at the derivative \( y' = 12x^2 - 3x^3 - x^3 \), which simplifies to \( y' = 12x^2 - 4x^3 \).
This derivative is essential for further analysis, as it gives us insights into the behavior of the function through critical evaluation of its rates of change.

Local Extremes

Local extremes of a function refer to the local maximum or minimum points where the function's value reaches a peak or a valley within a certain interval. To find these points, one must first identify where the derivative \( y' \) of the function equals zero or is undefined.
For our function, the derivative is \( y' = 4x^2(3-x) \). Setting \( y' = 0 \) identifies the critical points:

• \( 4x^2 = 0 \) gives \( x = 0 \)
• and \( 3-x = 0 \) gives \( x = 3 \)

Both \( x = 0 \) and \( x = 3 \) are critical points, which are potential local extremes. Evaluating the function at these points will reveal if they are indeed local maxima or minima, based on the changes in \( y \) as \( x \) crosses these points.

Inflection Points

Inflection points are locations on a curve where the curvature changes direction, signifying a shift in concavity from concave up to concave down, or vice versa. These points are found by analyzing the second derivative of the function.
The initial objective is to find \( y'' \), the second derivative. However, in some cases, further simplification of expressions from \( y' \) aids in this derivation. From \( y' = 4x^2(3-x) \), we identify the second derivative \( y'' \) by applying the product and chain rules repeatedly and eventually testing \( y'' = 0 \) or where \( y'' \) doesn't exist. This lets us determine potential inflection points.
After identifying these critical positions, analyzing the sign changes around these points in the second derivative confirms the nature and existence of an inflection point, aiding greatly in comprehending the shape and behavior of the graph.