Problem 12 - Calculus Early Transcendentals

Question

Find the absolute maximum and minimum values of \(f\) on the given closed interval, and state where those values occur. \(f(x)=\left(x^{2}+x\right)^{2 / 3} ;[-2,3]\)

Step-by-Step Solution

Answer

The absolute maximum is \( 12^{2/3} \) at \( x=3 \), and the minimum is 0 at \( x=0 \) and \( x=-1 \).

Step 1: Find the Critical Points

Find the derivative of the function to determine where the critical points occur. The function is given by \( f(x) = (x^2 + x)^{2/3} \). Set \( u = x^2 + x \), so \( f(x) = u^{2/3} \). Find the derivative \( f'(x) = \frac{d}{dx}(u^{2/3}) = \frac{2}{3}u^{-1/3} \cdot u' \), where \( u' = 2x + 1 \). Therefore, the derivative \( f'(x) = \frac{2}{3}(x^2+x)^{-1/3}(2x+1) \). Set this equal to zero to find critical points: \( \frac{2}{3}(x^2+x)^{-1/3}(2x+1) = 0 \) implies \( 2x+1 = 0 \), giving \( x = -\frac{1}{2} \). Check where the derivative is undefined to find \( x^2+x = 0 \), which gives additional critical points \( x = 0 \) and \( x = -1 \).

Step 2: Evaluate Function at Critical Points and Endpoints

Evaluate \( f(x) \) at the critical points and the endpoints of the interval. The endpoints are \( x = -2 \) and \( x = 3 \).Calculate:- \( f(-2) = ((-2)^2 + (-2))^{2/3} = (4-2)^{2/3} = 2^{2/3} \)- \( f(3) = (3^2 + 3)^{2/3} = (9+3)^{2/3} = 12^{2/3} \)- \( f\left(-\frac{1}{2}\right) = ((-\frac{1}{2})^2 - \frac{1}{2})^{2/3} = (\frac{1}{4} - \frac{1}{2})^{2/3} = \left(-\frac{1}{4}\right)^{2/3} \)- \( f(0) = (0^2 + 0)^{2/3} = 0 \)- \( f(-1) = ((-1)^2 + (-1))^{2/3} = (1-1)^{2/3} = 0 \)

Step 3: Compare Function Values

Evaluate the function values calculated in Step 2. Compare to find the absolute maximum and minimum on the closed interval:- \( f(-2) = 2^{2/3} \approx 1.5874 \)- \( f(3) = 12^{2/3} \approx 3.31 \)- \( f\left(-\frac{1}{2}\right) = \left(\frac{1}{4}\right)^{2/3} \approx 0.3969 \)- \( f(0) = 0 \)- \( f(-1) = 0 \)The absolute maximum value is \( 12^{2/3} \approx 3.31 \) at \( x = 3 \). The absolute minimum value is 0 at \( x = 0 \) and \( x = -1 \).

Key Concepts

Critical Points

Critical points play a fundamental role in calculus as they help us understand where a function changes direction. In mathematical terms, critical points are points where the derivative of a function is zero or undefined. These points are essential because they can indicate potential maximum or minimum values of the function.
To find critical points, follow these steps:

• Compute the derivative of the function, which gives us the rate of change at any point.
• Set this derivative equal to zero, since critical points can occur where the slope is zero (a turning point).
• Additionally, find where the derivative is undefined, as these points could also correspond to critical points.

In our exercise, we found the critical points of the function by setting the derivative of \( f(x) = (x^2 + x)^{2/3} \) to zero and also checking where the derivative is undefined. This process revealed critical points at \( x = -\frac{1}{2}, 0, \text{ and } -1 \). Identifying these points allows us to determine regions of change and possible extreme values.

Absolute Maximum and Minimum

Finding the absolute maximum and minimum values of a function on a closed interval involves comparing the function values at critical points and endpoints. The absolute maximum is the greatest value a function takes on over the entire interval, while the absolute minimum is the lowest value.
To determine these absolute extremum values:

• Evaluate the function at each critical point identified within the interval.
• Evaluate the function at the endpoints of the interval. These are important because the function may reach its extreme values at the end rather than just at critical points.
• Compare all these function values.

For the function \( f(x) = (x^2 + x)^{2/3} \), the critical points and endpoints that have been evaluated include \( x = -2, 3, -\frac{1}{2}, 0, \text{ and } -1 \). By comparing these function values, we concluded that the absolute maximum value is approximately \( 3.31 \) at \( x = 3 \), and the absolute minimum value is \( 0 \) at both \( x = 0 \) and \( x = -1 \). This comparison gives a clear picture of the function's behavior on the specified interval.

Derivative

The derivative of a function is a powerful tool in calculus that represents the rate of change or the slope of the function at any given point. Understanding how to calculate and interpret derivatives is crucial for finding critical points and determining the behavior of functions.
The derivative of a function \( f(x) \) can be computed using various techniques, among which is the power rule, chain rule, and product rule. In our exercise, finding the derivative of \( f(x) = (x^2 + x)^{2/3} \) involved using the chain rule:

• First, define \( u = x^2 + x \).
• Then differentiate using \( f(x) = u^{2/3} \).
• Compute \( f'(x) = \frac{2}{3}u^{-1/3} \cdot u' \).
• Calculate \( u' = 2x + 1 \), thus obtaining \( f'(x) = \frac{2}{3}(x^2+x)^{-1/3}(2x+1) \).

This derivative helps us determine where the slope is zero or undefined, thus locating critical points. Understanding derivatives allows us to explore deeper properties of functions, such as concavity and inflection points, making them an essential tool in calculus.