Problem 27 - Calculus with Concepts in Calculus

Question

Find all extreme values (if any) of the given function on the given interval. Determine at which numbers in the interval these values occur. $$ f(x)=-\sin \sqrt[3]{x} ;\left[-\pi^{3} / 27, \pi^{3} / 8\right] $$

Step-by-Step Solution

Answer

Maximum is \(\frac{\sqrt{3}}{2}\) at \(x = -\frac{\pi^3}{27}\), minimum is \(-1\) at \(x = \frac{\pi^3}{8}\).

Step 1: Identify the Critical Points

To find the critical points, we need to take the derivative of the function and set it equal to zero. The function is \( f(x) = -\sin \sqrt[3]{x} \). Differentiate using the chain rule: \( f'(x) = -\cos \sqrt[3]{x} \cdot \frac{1}{3}x^{-\frac{2}{3}} \). Set the derivative equal to zero to find critical points: \[- \cos \sqrt[3]{x} \cdot \frac{1}{3}x^{-\frac{2}{3}} = 0.\] Since the factor \( \frac{1}{3}x^{-\frac{2}{3}} \) cannot be zero, \( \cos \sqrt[3]{x} = 0 \). This holds true when \( \sqrt[3]{x} = \frac{\pi}{2} + k\pi \) for integers \( k \). Therefore, \( x = \left(\frac{\pi}{2} + k\pi\right)^3 \). Check if these values lie within the interval \( \left[-\frac{\pi^3}{27}, \frac{\pi^3}{8}\right] \).

Step 2: Evaluate the Interval Boundaries

Evaluate the function at the endpoints of the interval \( f(x) \) at \( x = -\frac{\pi^3}{27} \) and \( x = \frac{\pi^3}{8} \).Calculate \( f\left(-\frac{\pi^3}{27}\right) = -\sin\left(\sqrt[3]{-\frac{\pi^3}{27}}\right) = -\sin(-\frac{\pi}{3}) = \frac{\sqrt{3}}{2} \).Calculate \( f\left(\frac{\pi^3}{8}\right) = -\sin\left(\sqrt[3]{\frac{\pi^3}{8}}\right) = -\sin(\frac{\pi}{2}) = -1 \).

Step 3: Compare Values and Determine Extremes

The function values calculated at the critical point(s) within the interval and the boundary values lead us to compare:- Critical values computed potentially in Step 1 based on satisfying \( \cos \sqrt[3]{x} = 0 \) within the interval (no real solution for \( x \) inside this interval).- Endpoint evaluations: \( f\left(-\frac{\pi^3}{27}\right) = \frac{\sqrt{3}}{2} \) and \( f\left(\frac{\pi^3}{8}\right) = -1 \).The maximum value is \( \frac{\sqrt{3}}{2} \) at \( x = -\frac{\pi^3}{27} \), and the minimum is \(-1\) at \( x = \frac{\pi^3}{8} \).

Step 4: Confirm Function Behavior

Validate the absence of any change in sign of the derivative within the interval and confirm that no other critical points exist. Reaffirm: Since no other x-values within the domain makes the derivative zero, it supports the extremum evaluations on endpoints are definitive for this problem.

Key Concepts

Derivative

A key step in finding the extreme values of a function is to determine its derivative. The derivative of a function tells us how the function value changes as its input changes. For the given function, \( f(x) = -\sin \sqrt[3]{x} \), we need to use calculus and differentiate it. Differentiation involves various rules, and recognizing the right one to apply is crucial to solving problems.

For most algebraic functions, knowing their derivatives by heart or being comfortable with differentiation techniques is handy. Here, we employ the chain rule in our calculations.

Chain Rule

The chain rule is an essential technique in calculus, especially when dealing with compositions of functions. It helps us differentiate a composite function, where one function is inside another function.

In our expression, \( \sin \sqrt[3]{x} \), the sine function is applied to the cube root of \( x \). By using the chain rule, we can express the derivative as: \( f'(x) = -\cos \sqrt[3]{x} \cdot \frac{1}{3}x^{-\frac{2}{3}} \).

Here's how it works:

• First, differentiate the outer function: \( -\sin u \to -\cos u \),
• Then, multiply by the derivative of the inner function \( u = \sqrt[3]{x} \to \frac{1}{3}x^{-\frac{2}{3}} \).

Mastery of the chain rule allows you to tackle diverse problems involving nested functions.

Critical Points

Critical points are places where the derivative of a function is zero or undefined. These points help us determine where a function could reach its highest or lowest values on an interval.

For \( f'(x) \), the critical points are found by solving the equation \( -\cos \sqrt[3]{x} \cdot \frac{1}{3}x^{-\frac{2}{3}} = 0 \). Simplifying, we find \( \cos \sqrt[3]{x} = 0 \), which means \( \sqrt[3]{x} = \frac{\pi}{2} + k\pi \) for integers \( k \).

However, we need to ensure these value(s) fall within the interval \( \left[-\frac{\pi^3}{27}, \frac{\pi^3}{8}\right] \). This is crucial because critical points outside the interval are not relevant for finding extremes on a specified range.

Maximum and Minimum Values

Once critical points and endpoints are identified, evaluating the function at these points allows us to find the maximum and minimum values in a given interval. These are the so-called extreme values.

For this exercise, the endpoints of the interval are significant as critical values within it do not satisfy the conditions:

• The endpoint evaluation at \( x = -\frac{\pi^3}{27} \) yields \( f(x) = \frac{\sqrt{3}}{2} \).
• At \( x = \frac{\pi^3}{8} \), \( f(x) = -1 \).

These represent the boundary values of the function within the interval.

In the context of this specific function, the maximum value is \( \frac{\sqrt{3}}{2} \) and the minimum value is -1, demonstrating that endpoint evaluations are sometimes the key to finding extreme values.