Problem 11 - Calculus - AP Edition

Question

Locate the critical points and identify which critical points are stationary points. $$f(x)=\sqrt[3]{x^{2}-25}$$

Step-by-Step Solution

Answer

Stationary point: \(x = 0\); Non-stationary points: \(x = \pm 5\).

Step 1: Find the Derivative

To locate the critical points, we first need to find the derivative of the function using the chain rule. The function is \(f(x) = (x^2 - 25)^{1/3}\). Differentiating both sides with respect to \(x\) gives us:\[ f'(x) = \frac{1}{3}(x^2 - 25)^{-2/3} \cdot 2x = \frac{2x}{3(x^2 - 25)^{2/3}}.\]

Step 2: Locate Critical Points

Critical points occur where the derivative is zero or undefined. First, we set the derivative equal to zero:\[\frac{2x}{3(x^2 - 25)^{2/3}} = 0.\]The only solution to this equation is \(x = 0\), since the numerator \(2x = 0\) implies \(x = 0\).

Step 3: Check for Undefined Points

Next, we identify where the derivative is undefined. The derivative is undefined when the denominator is zero, i.e.,\[3(x^2 - 25)^{2/3} = 0.\]This occurs when \(x^2 - 25 = 0\), giving us \(x^2 = 25\), hence \(x = \pm 5\).

Step 4: Identify as Stationary or Non-stationary Points

A stationary point is where the derivative is zero. Since \(x = 0\) makes the derivative zero, it is a stationary point. However, \(x = \pm 5\) are non-stationary critical points as they make the derivative undefined.

Key Concepts

Chain Rule

The Chain Rule is a fundamental concept in calculus used to differentiate composite functions. When dealing with functions that are composed of other functions, the Chain Rule enables us to break the differentiation process into manageable steps. In the example of the function \(f(x) = \sqrt[3]{x^2 - 25}\), it consists of an inner function \(x^2 - 25\) and an outer function (the cube root). To find its derivative, we apply the Chain Rule, which requires finding the derivative of the outer function and multiplying it by the derivative of the inner function.

Here's a simple way to apply the Chain Rule:

• Differentiate the outer function while keeping the inner function unchanged.
• Multiply by the derivative of the inner function.

In the given solution, the derivative \(f'(x) = \frac{1}{3}(x^2 - 25)^{-2/3} \cdot 2x\) results from this rule. The term \(\frac{1}{3}(x^2 - 25)^{-2/3}\) comes from differentiating the cube root, and \(2x\) is the derivative of \(x^2 - 25\). Understanding the Chain Rule is essential for solving complex differentiation problems in calculus.

Stationary Points

Stationary Points are points on a graph where the slope of the tangent (derivative) is zero. It means that at these points, the function neither increases nor decreases, leading to horizontal tangents.

For the function in question, stationary points occur where the derivative \(f'(x)\) is zero. In this exercise, setting \(\frac{2x}{3(x^2 - 25)^{2/3}} = 0\) yields the solution \(x = 0\). Since the derivative is zero at this point, \(x = 0\) is identified as a stationary point. Finding these points is useful as they can be local maxima, minima, or saddle points of the function. They provide insights into the behavior of graphs and are critical in optimization problems.

Derivative

The Derivative of a function is a measure of how a function changes as its input changes. It represents the slope of the tangent line to the graph of the function at any given point.

In the exercise, the derivative \(f'(x)\) was calculated for the function \(f(x) = \sqrt[3]{x^2 - 25}\) using calculus rules like the Chain Rule. The resulting derivative \(\frac{2x}{3(x^2 - 25)^{2/3}}\) determines how the function's value changes as \(x\) changes.

Understanding derivatives is crucial because:

• They identify where the slope of the function changes.
• They help locate critical points which include stationary points.
• They tell us whether the function is increasing or decreasing at certain intervals.

In practice, derivatives are used for modeling, predicting changes, and optimizing across different fields from physics and engineering to economics and statistics.