Problem 32 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the absolute maximum value and the absolute minimum value, if any, of each function. $$ g(x)=x^{2}+2 x^{2 / 3} \text { on }[-2,2] $$

Step-by-Step Solution

Answer

The absolute minimum value of \(g(x) = x^2 + 2x^{\frac{2}{3}}\) on the interval \([-2, 2]\) is 0 at x = 0, and the absolute maximum value is 8 at x = -2 or x = 2.

Step 1: Find the first derivative of g(x)

We have the function \(g(x) = x^2 + 2x^{\frac{2}{3}}\). Using the power rule, find the first derivative \(g'(x)\): \(g'(x) = \frac{d}{dx}(x^2) + \frac{d}{dx}(2x^{\frac{2}{3}})\) \(g'(x) = 2x + \frac{4}{3}x^{-\frac{1}{3}}\)

Step 2: Find the critical points

Critical points are the points where the first derivative is either equal to 0 or undefined. So, we will set \(g'(x)\) equal to 0 and solve for x: \(0 = 2x + \frac{4}{3}x^{-\frac{1}{3}}\) Multiply both sides by \(3x^{\frac{1}{3}}\) to get rid of the fraction and the negative exponent: \(0 = 6x^{\frac{4}{3}} + 4x\) Factor out common term 2x: \(0 = 2x(3x^{\frac{1}{3}} + 2)\) Set each factor equal to 0: \(2x = 0 \rightarrow x = 0\) \(3x^{\frac{1}{3}} + 2 = 0 \rightarrow x^{\frac{1}{3}} = -\frac{2}{3}\) Since there is no real solution for the second equation (there is no real number whose cube root is negative), the only critical point is \(x = 0\).

Step 3: Find the function values at critical points and endpoints

Evaluate the function \(g(x)\) at the critical point x = 0 and the endpoints x = -2 and x = 2: At x = -2: \(g(-2) = (-2)^2 + 2(-2)^{\frac{2}{3}} = 4 + 2(2) = 8\) At x = 0: \(g(0) = 0^2 + 2(0)^{\frac{2}{3}} = 0 + 2(0) = 0\) At x = 2: \(g(2) = 2^2 + 2(2)^{\frac{2}{3}} = 4 + 2(2) = 8\)

Step 4: Determine the absolute maximum and minimum values

Compare the function values at the critical point and the endpoints: g(-2) = 8 g(0) = 0 g(2) = 8 The absolute minimum value of \(g(x)\) in the interval \([-2, 2]\) is 0 when x = 0. The absolute maximum value of \(g(x)\) is 8 when x = -2 or x = 2.

Key Concepts

Critical Points

Critical points are important in calculus optimization because they help identify where a function may have hills or valleys, which is essential when looking for maxima or minima. To locate critical points for the function \( g(x) = x^2 + 2x^{\frac{2}{3}} \), we start by finding the first derivative, denoted \( g'(x) \). This derivative shows how the function's output changes as \( x \) changes.

We calculate \( g'(x) \) to be \( 2x + \frac{4}{3}x^{-\frac{1}{3}} \). Critical points occur where this derivative is either zero or undefined. Setting \( g'(x) = 0 \) and solving for \( x \) allows us to find potential critical points:

• \( 2x = 0 \) gives us \( x = 0 \).
• The other factor \( 3x^{\frac{1}{3}} + 2 \) does not provide a real solution where it equals zero.

Thus, \( x = 0 \) is the only critical point for this function. It’s crucial to test this point since it helps locate where maximum or minimum values might occur on a given interval.

Endpoints Evaluation

Endpoints evaluation is essential, especially when the function is defined on a closed interval. This means observing how the function behaves at the very start and end of the interval, alongside any critical points. For our function \( g(x) = x^2 + 2x^{\frac{2}{3}} \) on the interval \([-2, 2]\), we need to evaluate \( g(x) \) at \( x = -2 \) and \( x = 2 \).

Why? Because sometimes, an absolute extreme value might occur at one of these endpoints, rather than at a critical point within the interval. It involves straightforward substitution of these \( x \) values into the original function:

• At \( x = -2 \), \( g(-2) = 8 \).
• At \( x = 2 \), \( g(2) = 8 \).

By carefully evaluating the function at these points, we can determine their importance in finding the absolute maximum and minimum values of the function.

Absolute Maximum and Minimum

The goal of calculus optimization often is to find the absolute highest and lowest values (maximum and minimum) that a function can achieve within a specified interval. After identifying critical points and evaluating endpoints, it’s time to compare and identify which of these provides the lowest and highest values.

In our function \( g(x) = x^2 + 2x^{\frac{2}{3}} \):

• At the critical point \( x = 0 \), \( g(0) = 0 \) which is the absolute minimum value.
• At the endpoints \( x = -2 \) and \( x = 2 \), both give \( g(-2) = 8 \) and \( g(2) = 8 \), representing the absolute maximum value.

This demonstrates that absolute extremes can sometimes occur at multiple points within the interval, and it emphasizes the importance of evaluating both critical points and endpoints. This comparison solidifies the understanding of where the function tops and dips, which is pivotal in optimization tasks.