Problem 70 - University Calculus: Early Transcendentals

Question

Find the critical points, domain endpoints, and extreme values (absolute and local) for each function. $$y=x^{2 / 3}\left(x^{2}-4\right)$$

Step-by-Step Solution

Answer

Critical points: x = -1, 0, 1. Local minima at x = -1, 0, 1, with values -3 and 0.

Step 1: Define the Function and Domain

The function we are working with is \( y = x^{2/3}(x^2 - 4) \). Since we have a cube root, this function is defined for all real numbers, as all numbers have real cube roots. Therefore, the domain of the function is \( (-\infty, \infty) \).

Step 2: Find the Derivative

To find the critical points, we first need the derivative of the function. Using the product rule, set \( u = x^{2/3} \) and \( v = x^2 - 4 \). The derivatives are \( u' = \frac{2}{3}x^{-1/3} \) and \( v' = 2x \). The derivative \( y' \) is given by:\[y' = u'v + uv' = \frac{2}{3}x^{-1/3}(x^2 - 4) + x^{2/3}\times 2x\]This simplifies to:\[y' = \frac{2}{3}(x^{-1/3})(x^2 - 4) + 2x^{5/3}\]

Step 3: Find Critical Points

Set the derivative \( y' \) equal to zero:\[\frac{2}{3}x^{-1/3}(x^2 - 4) + 2x^{5/3} = 0\]Factor out \( \frac{2}{3}x^{-1/3} \):\[\frac{2}{3}x^{-1/3}[(x^2 - 4) + 3x^2] = 0\]Simplify and solve:\[\frac{2}{3}x^{-1/3}(4x^2 - 4) = 0\]This gives \( 4x^2 - 4 = 0 \), resulting in \( x^2 = 1 \). Thus, critical points are \( x = 1 \) and \( x = -1 \). Also consider where the derivative is undefined, \( x = 0 \), as a critical point.

Step 4: Evaluate Function at Critical Points

Evaluate \( y \) at the critical points:\- At \( x = 1 \): \( y = 1^{2/3}(1^2 - 4) = -3 \)- At \( x = -1 \): \( y = (-1)^{2/3}((-1)^2 - 4) = -3 \)- At \( x = 0 \): \( y = 0^{2/3}(0 - 4) = 0 \)

Step 5: Determine Local and Absolute Extrema

Check if the critical points are minima, maxima, or neither by considering behavior around these points and using derivatives if needed.- \( x = 0 \): Since the derivative is undefined and the value is 0, this is a local minimum.- \( x = 1 \) and \( x = -1 \): Since these points produce the same value \(-3\) and the signs of \( y' \) change around them, they are local minima. With domain being infinite, check values at bounds to decide if these are also absolute minimum, which they are not necessarily in infinite domain. Therefore, the local minimum value is \(-3\) at \( x = 1 \) and \( x = -1 \), while local minimum at \( x = 0 \) is 0.

Key Concepts

Critical Points

Critical points are vital in understanding where a function reaches its peaks and troughs. To find critical points, you must identify where the derivative of the function is either zero or undefined. These points inform us about possible local extrema, such as local minima or maxima.

In our exercise, we set the derivative of the function to zero and solve for the variable, in this case, solving \[\frac{2}{3}x^{-1/3}((x^2 - 4) + 3x^2) = 0\] reveals the potential critical points. Additionally, examining where the derivative doesn't exist, such as at \( x = 0 \), provides further critical points.

Thus, through these calculations, we identified the critical points of the function to be \( x = 1 \), \( x = -1 \), and \( x = 0 \). These points warrant further investigation to determine their nature.

Derivative

The derivative of a function is a tool used to determine the rate of change or the slope at any point along the curve of the function. For this exercise, understanding the derivative is crucial in finding the critical points.

The product rule, a derivative rule used when dealing with products of functions, helps us to differentiate our function \( y = x^{2/3}(x^2 - 4) \).

By splitting into components:

• \( u = x^{2/3} \)
• \( v = x^2 - 4 \)

we differentiate each part individually:

• \( u' = \frac{2}{3}x^{-1/3} \)
• \( v' = 2x \)

Combining these using the product rule yields the derivative:\[y' = u'v + uv' = \frac{2}{3}x^{-1/3}(x^2 - 4) + x^{2/3} \times 2x\]

This simplifies to \( y' = \frac{2}{3}x^{-1/3}(x^2 - 4) + 2x^{5/3} \), giving a full picture of the function's behavior.

Local Extrema

Local extrema refer to the local maximum and minimum values of a function within a particular interval. These can be found by analyzing the critical points and the behavior of the derivative around these points.

In this problem, we identified the critical points \( x = 0 \), \( x = 1 \), and \( x = -1 \). By evaluating the function's derivative around these points, we find that \( x = 0 \) gives a local minimum since the derivative is undefined, and the value is zero at this point.

Both \( x = 1 \) and \( x = -1 \) also serve as local minima because the function values decrease to \(-3\) at these points, with sign changes in the derivative indicating different direction transitions. Understanding these local minima helps in sketching function graphs and predicting behavior within intervals.

Absolute Extrema

Absolute extrema are the highest and lowest values a function can achieve over its entire domain. Unlike local extrema, they consider the function values over the entire span rather than in small intervals.

In an infinite domain such as this one, as indicated by the function \( y = x^{2/3}(x^2 - 4) \), determining absolute extrema can be challenging. However, evaluating values at endpoints of finite intervals or critical points can assist in identifying these extrema.

In this example, while local minima are found at \( x = 0 \), \( x = -1 \), and \( x = 1 \), determining if these are absolute requires comparing with function behavior at large positive and negative limits of \( x \). Since our domain extends to infinity, extreme values aren't confined to finite numbers, and thus absolute extrema may not exist within the tested range. Nevertheless, noting significant function values aids in understanding potential extremities of the graph.