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Chapter 4: Problem 24

$$ \begin{array}{l}{\text { (a) Find the critical numbers of } f(x)=x^{4}(x-1)^{3} \text { . }} \\ {\text { (b) What does the Second Derivative Test tell you about the }} \\ {\text { behavior of } f \text { at these critical numbers? }} \\ {\text { (c) What does the First Derivative Test tell you? }}\end{array} $$

Short Answer

Expert verified
Critical numbers are 0, 1, and \( \frac{4}{7} \). The function's behavior varies across critical points and detailed tests determine minima/maxima.

Step by step solution

01

Find the first derivative

To find the critical numbers, we start by finding the first derivative of the function using the product rule. Given the function \( f(x) = x^4 (x-1)^3 \), let's differentiate it.Let \( u = x^4 \) and \( v = (x-1)^3 \), then \( f(x) = u \cdot v \).The product rule is \( (uv)' = u'v + uv' \).Find the derivatives: \( u' = 4x^3 \) and \( v' = 3(x-1)^2 \).So, \( f'(x) = 4x^3(x-1)^3 + x^4 \cdot 3(x-1)^2 \).Simplify to:\( f'(x) = x^3(x-1)^2 (4(x-1) + 3x) = x^3(x-1)^2 (7x - 4) \).
02

Find critical numbers

The critical numbers occur where \( f'(x) = 0 \) or where \( f'(x) \) is undefined. Since \( f'(x) \) is a polynomial, it is defined everywhere. Set the expression equal to zero:\[ x^3(x-1)^2(7x-4) = 0 \]Solve for \( x \):1. \( x^3 = 0 \Rightarrow x = 0 \).2. \( (x-1)^2 = 0 \Rightarrow x = 1 \).3. \( 7x - 4 = 0 \Rightarrow x = \frac{4}{7} \).The critical numbers are \( x = 0, \frac{4}{7}, \) and \( 1 \).
03

Find the second derivative

Next, we find the second derivative \( f''(x) \) to use the Second Derivative Test. Differentiate \( f'(x) = x^3(x-1)^2(7x-4) \) again using the product rule and chain rule:First simplify \( f'(x)=x^3 (x-1)^2 (7x - 4) \):1. Derivative of \( x^3 \) is \( 3x^2 \).2. Use the product rule on \((x-1)^2 (7x - 4)\):\( \, (x-1)^2 \cdot 7 = 14x(x-1) \) after using chain and product rules.Combine these into second derivative \( f''(x) \). This expression is complex, consider specific substitution to test critical points rather than explicit expression.
04

Second Derivative Test

Use the Second Derivative Test on each critical number to determine the behavior of the function, i.e., whether the point is a local minimum, maximum, or neither.1. For \( x = 0 \), find \( f''(0) \).2. For \( x = \frac{4}{7} \), find \( f''\left(\frac{4}{7}\right) \).3. For \( x = 1 \), find \( f''(1) \).Calculate and interpret results:- If \( f''(x) > 0 \), it indicates a local minimum.- If \( f''(x) < 0 \), it indicates a local maximum.- If \( f''(x) = 0 \), the test is inconclusive.
05

First Derivative Test

To understand the behavior at each critical number, apply the First Derivative Test. Evaluate \( f'(x) \) around each critical number to determine if the function is increasing or decreasing:Evaluate intervals around critical points:1. For \( x = 0 \), check \( f'(x) \) for \( x < 0 \) and \( x > 0 \).2. For \( x = \frac{4}{7} \), check around \( x < \frac{4}{7} \) and \( x > \frac{4}{7} \).3. For \( x = 1 \), check \( f'(x) \) for \( x < 1 \) and \( x > 1 \).Determine if each point is a local minimum, maximum, or neither by observing changes in monotonicity.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

First Derivative Test
The First Derivative Test is a method in calculus used to determine the local extrema of a function, such as local minima or maxima, by analyzing the sign changes of the first derivative around critical numbers. To perform this test, follow these steps:
  • Find the critical numbers by setting the first derivative of the function equal to zero or identifying where it is undefined. In our example, with the function \(f(x) = x^4(x-1)^3\), the critical numbers were found to be \(x = 0\), \(x = \frac{4}{7}\), and \(x = 1\).
  • Choose test points in intervals around each critical number. For example, if your critical point is \(x = \frac{4}{7}\), you might choose test values on either side, such as \(x < \frac{4}{7}\) and \(x > \frac{4}{7}\).
  • Evaluate the first derivative, \(f'(x)\), at these test points. Determine whether the function changes from increasing to decreasing or vice versa as you pass through the critical numbers.
  • If \(f'(x)\) changes from positive to negative, the critical number is a local maximum. If it changes from negative to positive, it indicates a local minimum. No sign change means the point may not be an extremum.
Using the First Derivative Test gives a simple but powerful insight into the behavior of functions around their critical numbers.
Second Derivative Test
The Second Derivative Test is another powerful tool we use in calculus to analyze the concavity of a function around its critical numbers, telling us more about the function's behavior. Here's how it works:
  • First, identify the critical numbers \(x = 0, \frac{4}{7}, \text{and } 1\) as mentioned in the problem.
  • Calculate the second derivative \(f''(x)\). For our function, this involves differentiating \(f'(x) = x^3(x-1)^2(7x-4)\) again, which you can do using the product and chain rules. While simplification may be complex, focus on evaluating specific critical points.
  • Evaluate \(f''(x)\) at your critical numbers.
  • Interpret the results:
    • If \(f''(x) > 0\), the function is concave up at that point, suggesting a local minimum.
    • If \(f''(x) < 0\), the function is concave down, indicating a local maximum.
    • If \(f''(x) = 0\), the test is inconclusive, and other methods, like the First Derivative Test, may be needed.
Overall, the Second Derivative Test gives a clear picture of the curvature at critical points, which is useful in understanding the nature of extrema.
Polynomial Functions
Polynomial functions are expressions that involve variables raised to whole number powers with constant coefficients, such as \(f(x) = x^4(x-1)^3\). Understanding their fundamental properties can be crucial:
  • Polynomial functions are smooth and continuous. They do not have breaks or sharp corners, which makes calculation of derivatives straightforward.
  • The degree of the polynomial is determined by the highest power of the variable. In our function, the degree is 7 (from \(x^4 \times x^3\)), suggesting potential for up to 6 turning points.
  • Critical points are solutions to the equation \(f'(x) = 0\). Since polynomials are defined everywhere, the first derivative is also defined everywhere, simplifying the process of finding these points.
  • By analyzing roots and behavior changes around critical points using derivatives, we understand the geometric nature of the polynomial.
These properties highlight the significance of polynomial functions in both theoretical and applied math, providing a foundational understanding of their behavior and applications.

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Most popular questions from this chapter

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