Problem 37 - Thomas Calculus in SI Units

Question

Find the function's absolute maximum and minimum values and say where they are assumed. $$f(x)=x^{4 / 3}, \quad-1 \leq x \leq 8$$

Step-by-Step Solution

Answer

Absolute maximum is 16 at \(x=8\), absolute minimum is 0 at \(x=0\).

Step 1: Find the critical points

To find the critical points, we first need to find the derivative of the function. The function is given by \(f(x)=x^{4/3}\). Let's calculate the derivative: \[f'(x) = \frac{4}{3}x^{-2/3}\]Set the derivative equal to zero to find critical points: \[\frac{4}{3}x^{-2/3} = 0\]Since the derivative will never be zero (as \(x\) must be a non-zero value to make the expression zero and exponent -2/3 results in undefined expression for zero), we must consider the places where derivative of the function does not exist. For this function, \(x=0\) causes \(f'(x)\) to be undefined. Therefore, \(x=0\) is a critical point.

Step 2: Evaluate endpoints and critical points

Now we evaluate the function \(f(x)=x^{4/3}\) at the endpoints and at any critical points to find the absolute max and min within the interval.Calculate \(f(x)\) at the endpoints and at \(x=0\):\[ f(-1) = (-1)^{4/3} = 1\]\[ f(0) = (0)^{4/3} = 0\]\[ f(8) = (8)^{4/3} = 16\]

Step 3: Determine the absolute maximum and minimum

Compare the values from step 2 to determine which is the absolute maximum and which is the absolute minimum.- The absolute maximum value is \(16\) at \(x=8\).- The absolute minimum value is \(0\) at \(x=0\).

Key Concepts

Absolute Maximum and Minimum

Finding the absolute maximum and minimum of a function means identifying the highest and lowest values that a function can take on a specified interval. For any continuous function over a closed interval

• The absolute maximum is the greatest value the function achieves on that interval.
• The absolute minimum is the smallest value of the function over the same period.

This is crucial for understanding the behavior of the function.
These points can occur at: - Endpoints of the interval - Critical points within the interval where the derivative may be zero or does not exist To locate these points, you evaluate the function at the critical points and the interval's endpoints, then compare the results. In the case of the function provided, we evaluated the critical point at x=0 and endpoints at x=-1 and x=8. It was found that the absolute maximum value is 16 at x=8, and the absolute minimum value is 0 at x=0.

Derivative of Functions

The derivative of a function is an essential tool in calculus. It helps identify how a function changes at any given point. It's like looking at how fast your car goes at each moment of a journey, a measure of change or speed if you will.
Calculating a derivative involves some basic rules, notably the power rule: if you have a function of the form \(f(x)=x^n\), then the derivative \(f'(x)=nx^{n-1}\).
This rule was applied to find the derivative of the function \(f(x)=x^{4/3}\), resulting in:\[ f'(x) = \frac{4}{3}x^{-2/3} \]

• Check where this derivative is equal to zero. If so, those points are considered as potential critical points.
• If the derivative does not exist, you must consider these points carefully.

In this example, the derivative never equals zero, but it is undefined at x=0, which makes x=0 a critical point.

Calculus Problem Solving

Solving calculus problems often involves following a step-by-step approach that consists of identifying relevant mathematical features of the function in question. For instance, finding absolute max and min involves calculating derivatives and evaluating at critical points and endpoints.
When faced with a function and an interval:

• Find the derivative to locate critical points.
• Evaluate the function at these points and at the endpoints of the interval.
• Compare these values to determine the absolute maxima and minima for the function over the interval.

This structured approach ensures that you consider all possible candidates for maximum and minimum values.
For the exercise provided, this means calculating the derivative, identifying critical points, and then evaluating each potential point by plugging back into the original function. This systematic approach to analysis ensures no potential extremum is overlooked.