Problem 94 - University Calculus: Early Transcendentals

Question

You will use a CAS to help find the absolute extrema of the given function over the specified closed interval. Perform the following steps. a. Plot the function over the interval to see its general behavior there. b. Find the interior points where \(f^{\prime}=0 .\) (In some exercises, you may have to use the numerical equation solver to approximate a solution.) You may want to plot \(f^{\prime}\) as well. c. Find the interior points where \(f^{\prime}\) does not exist. d. Evaluate the function at all points found in parts (b) and (c) and at the endpoints of the interval. e. Find the function's absolute extreme values on the interval and identify where they occur. $$f(x)=2+2 x-3 x^{2 / 3}, \quad[-1,10 / 3]$$

Step-by-Step Solution

Answer

The absolute maximum is 7 at \(x = -1\), and the absolute minimum is -8/3 at \(x = 10/3\).

Step 1: Plot the Function

Plot the function \(f(x) = 2 + 2x - 3x^{2/3}\) over the interval \([-1, 10/3]\) using a graphing tool or a Computer Algebra System (CAS). Look for high and low points visually to understand its behavior over the interval.

Step 2: Find Interior Critical Points Where Derivative is Zero

First, find the derivative of the function, \(f'(x) = 2 - 2x^{-1/3}\). Set \(f'(x) = 0\) and solve for \(x\): \(2 - 2x^{-1/3} = 0\) This simplifies to \(x^{-1/3} = 1\), hence \(x = 1\).

Step 3: Find Points Where Derivative Does Not Exist

The derivative \(f'(x) = 2 - 2x^{-1/3}\) does not exist where the expression is undefined. This occurs when the denominator of \(x^{-1/3}\) is zero, which happens at \(x = 0\). Thus, \(x = 0\) is another critical point.

Step 4: Evaluate the Function at Critical Points and Endpoints

Calculate \(f(x)\) at the critical points and the interval endpoints:- At \(x = -1\), calculate \(f(-1) = 2 + 2(-1) - 3(-1)^{2/3}\).- At \(x = 0\), calculate \(f(0) = 2 + 2(0) - 3(0)^{2/3}\).- At \(x = 1\), calculate \(f(1) = 2 + 2(1) - 3(1)^{2/3}\).- At \(x = \frac{10}{3}\), calculate \(f(\frac{10}{3}) = 2 + 2\frac{10}{3} - 3(\frac{10}{3})^{2/3}\).Evaluate to find the corresponding \(f(x)\) values.

Step 5: Determine Absolute Extreme Values

Compare the values from Step 4:- \(f(-1) = 7\)- \(f(0) = 2\)- \(f(1) = 1\)- \(f(\frac{10}{3}) = -\frac{8}{3}\)The absolute maximum is \(f(-1) = 7\) at \(x = -1\), and the absolute minimum is \(f(\frac{10}{3}) = -\frac{8}{3}\) at \(x = \frac{10}{3}\).

Key Concepts

Critical Points

Critical points of a function are where the derivative is zero or does not exist. These are the potential turning points where a function might change direction.

To find critical points, follow these steps:

• First, derive the function to get its derivative.
• Next, solve for when the derivative equals zero to find points where there might be a maximum or minimum.
• Identify where the derivative is undefined, as these can also be critical points.

In our exercise, the critical points were found at \(x = 1\) where the derivative is zero, and \(x = 0\), where the derivative is undefined.

Derivative

The derivative of a function gives you the rate of change or the slope of the function at any point. This is fundamental in calculus for determining how a function behaves.

To find a derivative, apply differentiation rules, such as the power rule or the chain rule. For example, to find the derivative of the given function \(f(x) = 2 + 2x - 3x^{2/3}\), we used differentiation to get:

\[f'(x) = 2 - 2x^{-1/3}\]

This derivative helps us understand where the function is increasing or decreasing, and thus where the critical points might be.

Extrema

Extrema refer to the maximum or minimum values a function can have, either within a certain interval (local) or on the entire function (absolute). These are essential in understanding the overall behavior of a function.

Extrema can be found by examining the critical points:

• Calculate the function values at these points.
• Compare them to find which are the highest or lowest.

In the exercise, we evaluated the function at critical points and interval endpoints to identify extrema, finding both local and absolute extrema.

Absolute Maximum and Minimum

The absolute maximum is the highest point over the entire interval, while the absolute minimum is the lowest. These are significant for problems requiring optimization or finding ultimate limits of a function over an interval.

To find these, evaluate the function at all critical points as well as at the boundaries of the interval. Compare these values:

The exercise showed the absolute maximum occurred at \(x = -1\), with \(f(x) = 7\), and the absolute minimum at \(x = \frac{10}{3}\), with \(f(x) = -\frac{8}{3}\).

By comparing function values, we can pinpoint these extremes, providing insights into the range of the function on the interval.