Problem 132 - Calculus Volume 1

Question

Find the local and absolute minima and maxima for the functions over \((-\infty, \infty)\). \(y=x^{3}(1-x)^{6}\)

Step-by-Step Solution

Answer

The absolute minimum is at \( x=0 \), and the absolute maximum is at \( x=\frac{1}{2} \).

Step 1: Expand the Function

First, let us expand the function \( y = x^{3}(1-x)^{6} \). Use the binomial theorem to expand \( (1-x)^{6} \). The expansion is \( 1 - 6x + 15x^2 - 20x^3 + 15x^4 - 6x^5 + x^6 \). Hence, the expanded function becomes \( y = x^3 (1 - 6x + 15x^2 - 20x^3 + 15x^4 - 6x^5 + x^6) \).

Step 2: Derive the Function

Next, find the first derivative of \( y \) with respect to \( x \). Differentiate each term in the expanded form: \( y = x^3(1 - 6x + 15x^2 - 20x^3 + 15x^4 - 6x^5 + x^6) \). The derivative is \( y' = 3x^2(1-x)^6 - 6x^3(1-x)^5 \).

Step 3: Simplify the Derivative

Factor the derivative: \( y' = x^2 (1-x)^5 (3 - 9x) \). Simplifying it gives \( y' = x^2 (1-x)^5 (-6x + 3) \).

Step 4: Set the Derivative Equal to Zero

To find the critical points, set the derivative equal to zero: \( x^2 (1-x)^5 (-6x + 3) = 0 \). Solve for \( x \). The critical points are \( x = 0, x = 1, x = \frac{1}{2} \).

Step 5: Determine the Nature of Critical Points

Use the second derivative test to identify if these points are local minima, maxima or points of inflection. Calculate the second derivative and evaluate it at \( x = 0, x = 1, x = \frac{1}{2} \). Find that \( x = 0 \) is a local minimum and \( x = \frac{1}{2} \) is a local maximum. \( x = 1 \) is an inflection point.

Step 6: Verify Absolute Extrema at Critical Points

Evaluate the function at the critical points. Calculate \( y(0), y(\frac{1}{2}), \) and \( y(1) \). The absolute minimum occurs at \( x = 0 \) with \( y(0) = 0 \) and the absolute maximum occurs at \( x = \frac{1}{2} \) with \( y(\frac{1}{2}) = \frac{1}{64} \).

Key Concepts

Critical Points

Critical points are the values of \( x \) where the derivative of a function is zero or undefined. These points are crucial in determining where the function changes behavior, such as going from increasing to decreasing, or vice versa. To find critical points, set the first derivative equal to zero and solve for \( x \).
For example, given a function \( y = x^3(1-x)^6 \), you first find its derivative and set it equal to zero:

• \( y' = x^2 (1-x)^5 (-6x + 3) = 0 \)

Solving this equation gives the critical points \( x = 0, x = 1, \) and \( x = \frac{1}{2} \). These points warrant further investigation to determine their nature using other methods like the Second Derivative Test.

Derivatives

Derivatives provide the rate at which a function is changing at any given point. They are foundational in calculus, used to find critical points, slopes of tangents, and much more.
To derive the function \( y = x^3(1-x)^6 \), expand it first using the binomial theorem, then apply the rules of differentiation to each term. The primary role of the first derivative \( y' \) is to find critical points.

• Derivatives can be computed using standard rules like power, product, and chain rules.
• Once you obtain \( y' \), set it equal to zero to find the critical points.

After finding \( y' = x^2 (1-x)^5 (-6x + 3) \), you can detect where a function begins to slope upwards or downwards.

Local Minima and Maxima

Local minima and maxima are points where the function

• reaches a valley (minimum) or
• a peak (maximum),

compared to the points immediately beside them. To discover these points, examine the critical points found using the first derivative. The notion is to check the function's behavior before and after these critical points.
Using the Second Derivative Test or other methods, you can classify the critical points. For instance, if the second derivative at a critical point is positive, \( f''(x) > 0 \), it suggests a local minimum. Conversely, negative \( f''(x) < 0 \) indicates a local maximum. Thus, by examining the second derivative at \( x = 0 \) and \( x = \frac{1}{2} \), it confirmed their roles as a local minimum and maximum, respectively.

Second Derivative Test

The Second Derivative Test is an effective way to classify critical points found with the first derivative. It involves calculating the second derivative \( f''(x) \) of a function and then evaluating this derivative at the critical points.
The simplest rules of the test are:

• If \( f''(x) > 0 \), the function is concave up at that point, suggesting 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 the point could be an inflection point.

In the example problem, evaluating the second derivative at the critical points \( x = 0, 1, \text{and } \frac{1}{2} \), revealed \( x = 0 \) as a local minimum, \( x = \frac{1}{2} \) as a local maximum, and \( x = 1 \) as an inflection point, enriching our understanding of the function's behavior on \((-\infty, \infty)\).