Problem 29 - Calculus Single Variable

Question

In Exercises 29-40, find each critical point \(c\) of the given function \(f\). Then use the First Derivative Test to determine whether \(f(c)\) is a local maximum value, a local minimum value, or neither. $$ f(x)=x(x+1)^{3 / 2},-1

Step-by-Step Solution

Answer

To determine the critical points and analyze their nature, find the derivative, set it to zero, solve for critical points, and use the First Derivative Test to assess each point.

Step 1: Differentiate the Function

To find the critical points, we first need to find the derivative of the function. Given that \( f(x) = x(x+1)^{3/2} \), we must apply the product rule: \[ (uv)' = u'v + uv' \]where \(u = x\) and \(v = (x+1)^{3/2}\). The derivative of \(u\) is \(u' = 1\), and for \(v\) we use the chain rule: \[ v' = \frac{3}{2}(x+1)^{1/2} \cdot 1 = \frac{3}{2}(x+1)^{1/2} \].Therefore, the derivative of the function is:\[ f'(x) = 1 \cdot (x+1)^{3/2} + x \cdot \frac{3}{2}(x+1)^{1/2} \].Simplifying gives: \[ f'(x) = (x+1)^{1/2} \left((x+1) + \frac{3}{2}x\right) = (x+1)^{1/2} \left(\frac{5}{2}x + 1\right) \].

Step 2: Identify the differentiation rules needed

Examine the function to determine which differentiation rules apply: power rule, product rule, quotient rule, chain rule, or special function derivatives.

Step 3: Apply the differentiation rules

Differentiate each term of the function systematically, applying the chain rule for composite functions.

Step 4: Simplify the derivative

Combine like terms, factor where appropriate, and write the derivative in its simplest form.

Step 5: State the final answer

Write the final derivative clearly.

Step 6: Conclude with the answer

To determine the critical points and analyze their nature, find the derivative, set it to zero, solve for critical points, and use the First Derivative Test to assess each point.

Key Concepts

First Derivative Test

The First Derivative Test is a method used in calculus to determine whether a function has a local maximum or minimum at a critical point. Critical points occur where the derivative of a function is zero or undefined. Once you find these points, the First Derivative Test guides you by analyzing the derivative's sign changes around these points.

Here's how it works:

• Find the critical points where the derivative is zero or does not exist.
• Check the sign of the derivative on either side of each critical point.
• If the derivative changes from positive to negative, the function has a local maximum at that point.
• If the derivative changes from negative to positive, the function has a local minimum.
• If there's no change in sign, the point is neither a maximum nor a minimum.

Using the First Derivative Test helps assess the behavior of functions and identify important features like local extrema.

product rule

The product rule is an essential tool in calculus for finding the derivative of a product of two functions. If you have two functions, say \(u(x)\) and \(v(x)\), the product rule states:

• \( (uv)' = u'v + uv' \)

This means you take the derivative of the first function and multiply it by the second function, then add the product of the first function and the derivative of the second function.

In our original exercise, the function given is \( f(x) = x(x+1)^{3/2} \). Here, we identified \(u = x\) and \(v = (x+1)^{3/2}\). By using the product rule, you combine the derivatives \(1\) and \(\frac{3}{2}(x+1)^{1/2}\) to find the derivative of the entire function. This calculation is a critical step towards finding critical points and applying the First Derivative Test.

chain rule

The chain rule is a fundamental principle in calculus for differentiating composite functions. It's especially useful when you have a function nested inside another. If you have an outer function \(f(x)\) and an inner function \(g(x)\), the chain rule formula is:

• \( (f(g(x)))' = f'(g(x)) \, g'(x) \)

This signifies taking the derivative of the outer function evaluated at the inner function, then multiplying by the derivative of the inner function.

In the context of the original exercise, applying the chain rule was necessary for the function \((x+1)^{3/2}\). The outer function here is \(f(u) = u^{3/2}\) and the inner function is \(g(x) = x+1\). Hence, differentiating \(f(u)\) with respect to \(u\) gives \(\frac{3}{2}u^{1/2}\), and multiplying this by the derivative of the inner function \((g(x)=1)\) completes the computation. This application ensures an accurate derivative, which is indispensable for identifying critical points.

derivative

Understanding derivatives is crucial in calculus. A derivative represents the rate of change of a function concerning one of its variables. You can think of a derivative as the slope of the tangent line to the function at a given point, providing insights on how the function behaves locally.

In our exercise, the process started with differentiating the given function \( f(x) = x(x+1)^{3/2} \). Finding the derivative enables the identification of critical points, as these occur where the derivative equals zero or is undefined. Understanding derivatives allow further analysis like using the First Derivative Test, informing if the critical points indicate local maxima or minima.

For complex functions, involving operations like the product or chain rule ensures a correct derivative calculation, which is foundational for any subsequent steps in solving problems involving rates of change and optimization.