Problem 53 - Single Variable Calculus

Question

In Exercises \(41-60\), find the absolute maximum and absolute minimum values, if any, of the function. $$ f(x)=x^{2 / 3}\left(x^{2}-4\right) \text { on }[-1,2] $$

Step-by-Step Solution

Answer

The absolute maximum value of the function \(f(x) = x^{\frac{2}{3}}(x^2 - 4)\) on the interval \([-1, 2]\) is 0, occurring at \(x = 0\) and \(x = 2\). The absolute minimum value is -3, occurring at \(x = -1\).

Step 1: Find the critical points

To find the critical points of the function, we need to find the first derivative and then solve for x when the first derivative is equal to 0 or undefined. The function is \(f(x) = x^{\frac{2}{3}}(x^2 - 4)\). Let's first find the derivative: \(f'(x) = \frac{d}{dx}(x^{\frac{2}{3}}(x^2 - 4))\) Using the product rule: \((uv)' = u'v + uv'\) Let \(u = x^{\frac{2}{3}}\) and \(v = (x^2 - 4)\). Now, find the derivatives of \(u\) and \(v\): \(u' = \frac{2}{3}x^{\frac{-1}{3}}\) \(v' = 2x\) Now, apply the product rule: \(f'(x) = (\frac{2}{3}x^{\frac{-1}{3}})(x^2 - 4) + (x^{\frac{2}{3}})(2x)\) Now, find the critical points by setting \(f'(x)\) to 0 or undefined: \(0 = (\frac{2}{3}x^{\frac{-1}{3}})(x^2 - 4) + (x^{\frac{2}{3}})(2x)\) This equation simplifies to: \(0 = -\frac{2}{3}x^{\frac{2}{3}}(x^2 - 4) + 2x^{\frac{5}{3}}\) Solve the above equation for the critical points.

Step 2: Evaluate the function at critical points and endpoints

Now, we need to find and evaluate the function at the critical points and endpoints of the interval, \([-1, 2]\). From the previous step, we found one critical point at \(x = 0\). Now, evaluate the function at the critical point and endpoints: \(f(-1) = (-1)^{\frac{2}{3}}((-1)^2 - 4) = 1(-3) = -3\) \(f(0) = (0)^{\frac{2}{3}}((0)^2 - 4) = 0\) \(f(2) = (2)^{\frac{2}{3}}((2)^2 - 4) = 4^{\frac{2}{3}}(0) = 0\)

Step 3: Compare the function values to determine the absolute maximum and minimum

We have evaluated the function at the critical point and endpoints, and now we need to compare the function values to determine the absolute maximum and minimum: \(f(-1) = -3\) \(f(0) = 0\) \(f(2) = 0\) The absolute maximum value is 0, which occurs at \(x = 0\) and \(x = 2\). The absolute minimum value is -3, which occurs at \(x = -1\).

Key Concepts

Absolute Maximum and Minimum

Finding the absolute maximum and minimum values of a function is a critical technique in calculus, often necessary in real-world problem-solving. Absolute extremum occurs when we find the highest or lowest point over the entire range of a function. To determine these values, we need to evaluate the function not only at its critical points but also at the endpoints of the given interval.

• Begin by identifying the relevant interval for which you are seeking the maximum and minimum; in this exercise, it's a [-1, 2].
• Use derivative calculations to pinpoint critical points within the interval, as these are potential locations for extremum.
• Assess the function's value at these critical points and endpoints to identify the peaks and valleys.

In our example with the function \(f(x) = x^{2/3}(x^2 - 4)\), evaluating at these points allowed us to determine that the absolute maximum value is 0 and the absolute minimum value is -3.

Function Derivatives

Derivatives are fundamental in finding critical points of a function. A derivative gives us important information about the slope of a function at any point and helps us determine where a function is increasing or decreasing.

For our function, \(f(x)=x^{2/3}(x^{2}-4)\), we apply the product rule to find the derivative. The product rule is defined as \((uv)' = u'v + uv'\). Here, if you view \(x^{2/3}\) as \(u\) and \((x^2 -4)\) as \(v\), the derivative process involves:

• Finding \(u' = \frac{2}{3}x^{-1/3}\)
• Finding \(v' = 2x\)
• Applying the product rule resulting in: \(f'(x) = \left(\frac{2}{3}x^{-1/3}\right)(x^{2}-4) + (x^{2/3})(2x)\).

With the derivative in hand, critical points are found by solving \(f'(x) = 0\) or determining when it's undefined. These calculations revealed a critical point at \(x = 0\), which needed further evaluation.

Interval Evaluation

The principle of evaluating a function over a given interval is to identify where it reaches its extremum points. This is achieved by evaluating the endpoints along with any identified critical points within that interval. For the given function, \([-1, 2]\) is the interval we're examining.

Steps to evaluate:

• Identify critical points within the interval using the derivative to find where it equals zero or is undefined.
• Assess the function at these critical points and at the interval's endpoints. In our exercise, this resulted in evaluations at \(x = -1\), \(x = 0\), and \(x = 2\).
• Compare these values to determine where the function achieves its maximum and minimum values.

Evaluations showed function values of \(-3, 0,\) and \(0\) at the identified points, allowing us to conclude the absolute minimum is \(-3\) and the absolute maximum is \(0\). This process thoroughly checks both the interval's internal behavior and its boundary conditions, ensuring no potential extremum is overlooked.