Problem 52 - Thomas Calculus in SI Units

Question

Find the extreme values (absolute and local) of the function over its natural domain, and where they occur. $$y=x^{3}(x-5)^{2}$$

Step-by-Step Solution

Answer

The function has a local minimum at \( x = 3 \). No absolute extrema on \((-\infty, \infty)\).

Step 1: Understand the function and its domain

The given function is \( y = x^3 (x-5)^2 \). The domain of this polynomial function is all real numbers, i.e., \( (-\infty, \infty) \). There are no restrictions like division by zero or square roots of negative numbers.

Step 2: Find the first derivative

To find the extreme values, we need to determine the critical points by setting the first derivative to zero. The function is \( y = x^3 (x-5)^2 \). Apply the product rule to differentiate:The product rule states \( (u \cdot v)' = u'v + uv' \), where \( u = x^3 \) and \( v = (x-5)^2 \).We find: \( u' = 3x^2 \) and \( v' = 2(x-5) \).Thus, the derivative is: \[ y' = 3x^2(x-5)^2 + x^3 \cdot 2(x-5) = 3x^2(x-5)^2 + 2x^3(x-5) \].Simplify the expression to find:\[ y' = x^2 (x-5) (5x - 15) \].

Step 3: Solve for critical points

Set the derivative equal to zero and solve for \( x \):\[ x^2 (x-5) (5x - 15) = 0 \].This gives three factors: 1. \( x^2 = 0 \) thus \( x = 0 \), 2. \( x-5 = 0 \) thus \( x = 5 \),3. \( 5x - 15 = 0 \) simplifying gives \( x = 3 \).Thus, the critical points are \( x = 0, 3, 5 \).

Step 4: Use the first derivative test for local extrema

Evaluate the intervals created by the critical points using the first derivative test:1. Test intervals: \((-\infty, 0)\), \((0, 3)\), \((3, 5)\), \((5, \infty)\).2. Choose test points: for example \(-1, 1, 4, 6\).3. Check the sign of \( y' \) in these intervals:- For \( x = -1 \), \( y' > 0 \).- For \( x = 1 \), \( y' < 0 \).- For \( x = 4 \), \( y' > 0 \).- For \( x = 6 \), \( y' > 0 \).Thus, \( y \) decreases on \((0, 3)\) and increases on \((3, 5)\), and is increasing on other intervals.

Step 5: Classify the critical points

Based on the first derivative test:- At \( x = 0 \) and \( x = 5 \): \( y' \) does not change sign, thus these may not be local extremum points.- At \( x = 3 \): \( y' \) changes from negative to positive, indicating a local minimum.

Step 6: Identify absolute extrema

Check the function's behavior as \( x \to \pm\infty \). As a cubic polynomial, \( y \to \infty \) as \( x \to \pm\infty \). Since there are no highest or lowest values overall, we do not have absolute extrema.

Key Concepts

Critical Points

Critical points are where a function’s derivative either is zero or doesn't exist. Finding these points is essential because they are where a function can have a relative maximum, a relative minimum, or even a stationary point. In our given problem, we have determined that the critical points occur where the derivative, \( y' \), is zero. By setting the derivative \( y' = x^2 (x-5)(5x - 15) \) equal to zero, we find critical points at \( x = 0, 3, \) and \( 5 \). Each of these values makes at least one factor of the derivative zero. It's important to identify these points because, at these specific \( x \)-values, the function's slope changes direction or stagnates, which is key to locating local extrema.

First Derivative Test

The First Derivative Test is a handy tool for identifying local maxima and minima by examining the behavior of the derivative before and after the critical points. After we've found the critical points, the next step is to pick test points in intervals around these points to see how the sign of \( y' \) changes. For example, by evaluating \( y' \) at test points like \(-1, 1, 4, \) and \( 6 \) for our intervals, we observe that:

• \( y' > 0 \) when \( x = -1 \), meaning \( y \) is increasing in this interval.
• \( y' < 0 \) when \( x = 1 \), which shows \( y \) is decreasing in the interval \((0, 3)\).
• \( y' > 0 \) when \( x = 4 \), indicating \( y \) increases in the interval \((3, 5)\).
• \( y' > 0 \) when \( x = 6 \).

These changes in sign at \( x = 3 \) indicate a local minimum because the slope switches from decreasing (negative \( y' \)) to increasing (positive \( y' \)). This method helps robustly identify the nature of critical points as local extrema.

Polynomial Functions

Polynomial functions are expressions made up of variables raised to whole number powers and their coefficients. Our exercise deals with the polynomial function \( y = x^3 (x-5)^2 \), a polynomial of degree 5, which signifies its highest power term when expanded will involve \( x^5 \). Such functions are known for their smooth curves, which are continuous and have derivatives at all points within their domain. This makes polynomials particularly suited to derivative tests needed for finding extrema. They have complicated behavior as they reach high degrees: greater twisting and turning in the graph as seen with \( y = x^3 (x - 5)^2 \) which can lead to complex scenarios involving multiple extremums.

Local Extrema

Local extrema refer to the highest or lowest points within a specific interval in a function, rather than the entire domain. In simpler terms, they are local peaks or valleys. In solving for these in our exercise for the function \( y = x^3(x-5)^2 \), using the First Derivative Test, we found a local minimum at \( x = 3 \), as the first derivative \( y' \) changes from negative to positive at this point. This test does not show any local maxima within our critical points, thus indicating that peaks or valleys might occur at endpoints in other contexts of boundary-restricted domains. In our polynomial, it continues indefinitely in both directions, which is why absolute extrema couldn’t be verified. Understanding local extrema helps in understanding the overall shape and behavior of the graph.