Problem 30 - Calculus Early Transcendentals

Question

Find the intervals on which \(f\) is increasing and decreasing. $$f(x)=x^{2} \sqrt{9-x^{2}} \text { on }(-3,3)$$

Step-by-Step Solution

Answer

Question: Determine the intervals where the given function, \(f(x) = x^2 \sqrt{9 - x^2}\), is increasing or decreasing. Answer: The function is increasing on the intervals \((-3, -\sqrt{6})\) and \((0, +\sqrt{6})\), and decreasing on the intervals \((-\sqrt{6}, 0)\) and \((+\sqrt{6}, 3)\).

Step 1: Calculate the derivative, f'(x)

To find the derivative of this composite function, apply the product rule and the chain rule. The product rule states that \((uv)' = u'v + uv'\), and the chain rule states that if \(y = g(u)\) and \(u = h(x)\), then \(y' = g'(u) \cdot h'(x)\) (derivative of the outer function times the derivative of the inner function). \(f(x) = x^2 (9 - x^2)^{1/2}\) Putting the \(\sqrt{9-x^2}\) as power function. Applying the product rule: \(f'(x) = (x^2)'\cdot(9-x^2)^{1/2} + x^2\cdot[(9-x^2)^{1/2}]'\) NowFind the derivative of each term separately: \((x^2)' = 2x \) Use chain rule on \((9-x^2)^{1/2}\) \([(9-x^2)^{1/2}]' = (\frac{1}{2})(9-x^2)^{-1/2}\cdot(-2x)\) Now, substitute each derivative back into the product rule equation: \(f'(x) = 2x (9 - x^2)^{1/2} + x^2(\frac{-x}{(9-x^2)^{1/2}})\)

Step 2: Identify the critical points

To find the critical points, set \(f'(x)\) equal to zero and solve for \(x\): \(2x (9 - x^2)^{1/2} + x^2(\frac{-x}{(9-x^2)^{1/2}}) = 0\) This fraction expression can be simplified to: \(2x(9 - x^2)^{1/2} - x^3(9-x^2)^{-1/2} = 0\) To solve for \(x\), we can multiply both sides by \((9 - x^2)^{1/2}\): \(2x(9 - x^2) - x^3 = 0\) Factor out x: \(x(2(9 - x^2) - x^2) = 0\) So, \(x = 0\) is one of the critical points. Now solve for the other critical point: \(2(9 - x^2) - x^2 = 0\) Solving this equation, you get \(x^2 = 6\). Thus, the critical points are \(x = 0\) and \(x = \pm\sqrt{6}\).

Step 3: Determine the intervals where f'(x) > 0 and f'(x) < 0

Create a number line containing each critical point \((x=-\sqrt{6}, 0, +\sqrt{6})\). Select a test point within each interval and substitute it into \(f'(x)\). The intervals are: 1. \((-3, -\sqrt{6})\) - Test point: \(x = -2\) 2. \((-\sqrt{6}, 0)\) - Test point: \(x = -1\) 3. \((0, +\sqrt{6})\) - Test point: \(x = 1\) 4. \((+\sqrt{6}, 3)\) - Test point: \(x = 2\) Substitute the test points into \(f'(x)\) and check if it is positive or negative: 1. \(f'(-2) > 0\) - Increasing 2. \(f'(-1) < 0\) - Decreasing 3. \(f'(1) > 0\) - Increasing 4. \(f'(2) < 0\) - Decreasing Therefore, the intervals for increasing are \((-3, -\sqrt{6})\) and \((0, +\sqrt{6})\), and the intervals for decreasing are \((-\sqrt{6}, 0)\) and \((+\sqrt{6}, 3)\).

Key Concepts

Product Rule

When dealing with the derivative of a product of two functions, we use a principle called the product rule. This rule is essential because when two functions are multiplied, their derivatives are not simply the derivative of the first times the derivative of the second.
Instead, the product rule states that the derivative of a product is:

• \((uv)' = u'v + uv'\)

Here, \(u\) and \(v\) are functions of \(x\).
This means you first derive \(u\) and multiply it by \(v\), then add \(u\) times the derivative of \(v\).
It's important to apply this rule step by step and calculate each term separately. In our example, \(f(x) = x^2 (9-x^2)^{1/2}\), we started by letting \(u = x^2\) and \(v = (9-x^2)^{1/2}\) to apply the product rule correctly.
This systematic approach allows us to accurately find the derivative of more complex functions that are built from simpler parts.

Chain Rule

The chain rule is an indispensable tool when you have a function that is composed of other functions, often seen in nested or complex expressions. Simply put, the chain rule states how to differentiate a function that has a function inside it (or an outer and an inner function).
The formula is:

• \(y' = g'(u) \cdot h'(x)\)

In this scenario, \(y = g(u)\), where \(g\) is a function of \(u\), and \(u = h(x)\), with \(h\) being a function of \(x\).
This rule tells us that to find the derivative \(y'\) of \(y\) with respect to \(x\), we should:

• First, derive the outer function \(g\) with respect to its variable \(u\).
• Then, multiply it by the derivative of the inner function \(h\) with respect to \(x\).

In our problem, applying the chain rule was necessary for differentiating \((9-x^2)^{1/2}\). By treating \(9-x^2\) as the inner function and its square root as the outer function, you can correctly apply the chain rule to find the derivative.

Critical Points

Critical points are vital in understanding the behavior of functions because they help determine where a function changes direction, specifically where it shifts from increasing to decreasing or vice versa.
To find critical points, you first need to compute the derivative \(f'(x)\) of the function \(f(x)\) and set it equal to zero:

• \(f'(x) = 0\)

This process identifies points where the slope of the tangent is zero, indicating potential peaks, troughs, or flat points.
In the example, once \(f'(x)\) is calculated using the product and chain rules, setting it to zero reveals critical points at \(x = 0\) and \(x = \pm\sqrt{6}\).
Analyzing these points along with test intervals helps to clarify where the function is increasing or decreasing. Understanding critical points is essential for graphing functions and solving real-world problems where rate changes are significant.