Problem 37 - Thomas Calculus

Question

In Exercises \(37-40\) , find the function's absolute maximum and minimum values and say where they occur. $$ f(x)=x^{4 / 3},-1 \leq x \leq 8 $$

Step-by-Step Solution

Answer

The absolute minimum is 0 at x=0; the absolute maximum is 16 at x=8.

Step 1: Identify the Function and Interval

The function given is \( f(x) = x^{4/3} \) and it is defined on the closed interval \([-1, 8]\). Our task is to find the absolute maximum and minimum values of this function over this interval.

Step 2: Find Critical Points

To find critical points, we first need to compute the derivative of \( f(x) \). Using the power rule for derivatives, the derivative is \( f'(x) = \frac{4}{3}x^{1/3} \). Set \( f'(x) = 0 \) to locate critical points: \( \frac{4}{3}x^{1/3} = 0 \), which simplifies to \( x=0 \) as the only critical point.

Step 3: Evaluate Endpoints and Critical Points

Evaluate the function \( f(x) \) at the endpoints of the interval and at the critical point: - At \( x = -1 \), \( f(-1) = (-1)^{4/3} = 1 \).- At \( x = 0 \), \( f(0) = 0^{4/3} = 0 \).- At \( x = 8 \), \( f(8) = 8^{4/3} = 16 \).

Step 4: Determine Absolute Extrema

Compare the values obtained:- \( f(-1) = 1 \)- \( f(0) = 0 \)- \( f(8) = 16 \)The absolute minimum value is \( 0 \) at \( x=0 \) and the absolute maximum value is \( 16 \) at \( x=8 \).

Key Concepts

Absolute Maximum

The absolute maximum of a function is the highest point over its entire given domain. It is essential to keep in mind that the function might reach this point at different places, either at a critical point or at the endpoints of the interval. In this exercise, we evaluated the function

• at the endpoints
• at critical points

of the interval to determine where the absolute maximum occurs.

For the function \(f(x) = x^{4/3}\) over the interval \([-1, 8]\), the absolute maximum was found to be \(16\) at \(x = 8\). This resulted from evaluating \( f(x) \) at \(x=8\), as this value was greater than the values at other points checked. Understanding how to find an absolute maximum includes checking these strategic points thoroughly.

Absolute Minimum

The absolute minimum of a function represents the lowest point over a prescribed domain. It's where the function attains its smallest output value. To find this, one must evaluate the function similar to finding an absolute maximum: at both critical points and the edges of an interval.

In this exercise, for the function \(f(x) = x^{4/3}\) within the interval \([-1, 8]\), the absolute minimum was \(0\) at \(x = 0\). The calculation proved this by evaluating \( f(x) \) at \(x=0\), revealing that no other evaluated point yielded a smaller value. This step is crucial in optimization problems where minimizing a function's output is desired.

Critical Points

Critical points are specific values of \(x\) where the first derivative of a function is either zero or undefined. These points are necessary for locating possible locations of local maximums, minimums, or points of inflection.

In our problem, we derived the first derivative of \(f(x) = x^{4/3}\), which is \(f'(x) = \frac{4}{3}x^{1/3}\). Setting this derivative to zero, we found that \( x = 0 \) is a critical point. This discovery is key because evaluating the function at these critical points alongside endpoints tells us about absolute extrema. Remember that these points provide critical insights into a function's overall behavior and solutions to optimization problems.

Derivative

The derivative of a function gives us its rate of change and is instrumental in determining critical points. Using the power rule, we computed the derivative for our function \(f(x) = x^{4/3}\) to find \(f'(x) = \frac{4}{3}x^{1/3}\).

• The derivative helps identify where the function's slope might level out (critical points).
• It also guides where the function will rise or fall.

In calculus, derivatives are fundamental because they offer insights into function's behavior: increasing or decreasing, convex or concave. Studying a derivative's sign and value offers crucial clues in broader analyses, such as finding maximums and minimums on an interval.

Interval Analysis

Interval analysis involves studying a function over a particular range. In our task, we analyzed the function \(f(x) = x^{4/3}\) over the closed interval \([-1, 8]\).

• This meant evaluating and comparing the function at specific points: endpoints and critical points.
• Interval analysis helps in determining how a function behaves across a selected span.

Through this process, one might discover increasing or decreasing trends and find absolute extrema. Given the comparison of function values, we concluded our earlier findings for maximum and minimum values within this interval. Understanding an interval's influence on a function guides decision-making and predictions across various scenarios in calculus applications.