Problem 28 - Thomas Calculus

Question

In Exercises \(21-36,\) find the absolute maximum and minimum values of each function on the given interval. Then graph the function. Identify the points on the graph where the absolute extrema occur, and include their coordinates. $$ h(x)=-3 x^{2 / 3}, \quad-1 \leq x \leq 1 $$

Step-by-Step Solution

Answer

Absolute max: 0 at \(x = 0\); absolute min: -3 at \(x = -1\) and \(x = 1\).

Step 1: Identify the Function and Interval

We are given the function \(h(x) = -3x^{2/3}\) and the interval \(-1 \leq x \leq 1\). Our goal is to find the absolute maximum and minimum values of this function within the specified interval.

Step 2: Find the Derivative

To locate critical points, we first find the derivative of the function. Using the power rule for differentiation: \(h'(x) = -3 \cdot \frac{2}{3}x^{-1/3} = -2x^{-1/3}\).

Step 3: Find Critical Points

Critical points occur where the derivative is zero or undefined. In this case, \(h'(x) = -2x^{-1/3}\) is undefined at \(x = 0\). Therefore, \(x = 0\) is a critical point to consider within our interval.

Step 4: Evaluate the Function at Critical Points and Endpoints

We need to evaluate \(h(x)\) at the critical point and at the endpoints of the interval: - Endpoint \(x = -1\): \(h(-1) = -3(-1)^{2/3} = -3\)- Critical Point \(x = 0\): \(h(0) = -3(0)^{2/3} = 0\)- Endpoint \(x = 1\): \(h(1) = -3(1)^{2/3} = -3\)

Step 5: Determine the Absolute Extrema

Comparing the values obtained:- \(h(-1) = -3\)- \(h(0) = 0\)- \(h(1) = -3\)The absolute maximum value is \(0\) at \(x = 0\), and the absolute minimum value is \(-3\) at both \(x = -1\) and \(x = 1\).

Step 6: Graph the Function and Mark Extrema

Graph the function \(h(x) = -3x^{2/3}\) over the interval \(-1 \leq x \leq 1\). Mark the absolute maximum at \((0, 0)\) and the absolute minimums at \((-1, -3)\) and \((1, -3)\).

Key Concepts

Absolute Maximum

An absolute maximum of a function is the highest point on the graph within the given interval. It represents the largest y-value that the function reaches in that range. Finding the absolute maximum involves evaluating the function not only at critical points but also at the endpoints of the interval. In our exercise, the function \(h(x)=-3x^{2/3}\) has an absolute maximum value at \(x=0\), where the function value is \(h(0)=0\). It's important to check endpoints and critical points because the absolute maximum can occur in these places.

Absolute Minimum

The absolute minimum is the lowest point a function reaches within a certain interval. It is the smallest y-value within that range. Just like with the absolute maximum, determining the absolute minimum requires checking the function at both endpoints and critical points. In the case of our function \(h(x) = -3x^{2/3}\), the absolute minimum value is found at both endpoints, \(x=-1\) and \(x=1\), with a value of \(-3\). Keep in mind that functions can have multiple points where absolute minimums occur, especially when the graph is symmetric around the y-axis.

Critical Points

Critical points of a function are where the first derivative is either zero or undefined. These points are important because they can indicate potential maximums or minimums. To find them, first take the derivative of the function. For \(h(x) = -3x^{2/3}\), the derivative is \(h'(x) = -2x^{-1/3}\). The derivative is undefined at \(x = 0\) because dividing by zero is undefined, making \(x = 0\) a critical point. Critical points are candidates for absolute extrema but are not guaranteed to be actual extremes without further evaluation.

Derivative

The derivative of a function helps us understand how that function changes. It is a mathematical tool used to calculate the slope of the tangent line to the function at any given point. For the function \(h(x) = -3x^{2/3}\), the derivative \(-2x^{-1/3}\) tells us the rate of change at different x-values. This information is essential to identify critical points, as they occur when the derivative is zero or undefined. Differentiation, which finds the derivative, often uses rules like the power rule, product rule, or chain rule depending on the function's composition.

Graphing Functions

Graphing a function provides visual insight into its behavior over a range of values. For \(h(x) = -3x^{2/3}\) on the interval \(-1 \leq x \leq 1\), plotting points helps to see where extrema occur. By graphing, you can identify key points such as the absolute maximum at \((0, 0)\) and absolute minimums at \((-1, -3)\) and \((1, -3)\). The shape and direction of the graph can also indicate the nature of critical points and how the function behaves as x approaches different values within the interval. This visual representation supports analytical findings and can often make complex relationships more comprehensible.