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Find all critical values, the largest open intervals on which \(f\) is increasing, the largest open intervals on which \(f\) is decreasing, and all relative maxima and minima. Sketch a rough graph of \(f\). In Exercises 37 through 42, assume that the constants \(a\) and \(b\) are positive. \(f(x)=5+20 x-x^{5}\)

Short Answer

Expert verified
The critical points are \( x = \pm \sqrt{2} \). Increasing on \((-\sqrt{2}, \sqrt{2})\), decreasing on \((-\infty, -\sqrt{2})\) and \((\sqrt{2}, \infty)\). Relative max at \( x = \sqrt{2} \), min at \( x = -\sqrt{2} \).

Step by step solution

01

Find the first derivative

To find the critical values where the function's slope is zero, we first need to compute the first derivative of the function \( f(x) = 5 + 20x - x^5 \). The derivative is obtained using the power rule. Thus, \[ f'(x) = 20 - 5x^4. \]
02

Solve for critical points

Set the first derivative equal to zero to find the critical points:\[20 - 5x^4 = 0.\]Solve this equation for \( x \):\[5x^4 = 20 \x^4 = 4 \x = \pm \sqrt[4]{4} = \pm \sqrt{2}.\] Thus, the critical points are \( x = \sqrt{2} \) and \( x = -\sqrt{2} \).
03

Determine intervals of increase and decrease

To determine where the function is increasing or decreasing, we will test the sign of \( f'(x) = 20 - 5x^4 \) in the intervals formed by the critical points: \((-\infty, -\sqrt{2})\), \((-\sqrt{2}, \sqrt{2})\), and \((\sqrt{2}, \infty)\).Choose test points in each interval:- For \( x = -2 \) (in \((-\infty, -\sqrt{2})\)), \( f'(-2) = 20 - 5(-2)^4 = 20 - 80 = -60 \), indicating \( f \) is decreasing.- For \( x = 0 \) (in \((-\sqrt{2}, \sqrt{2})\)), \( f'(0) = 20 - 5(0)^4 = 20 \), indicating \( f \) is increasing.- For \( x = 2 \) (in \((\sqrt{2}, \infty)\)), \( f'(2) = 20 - 5(2)^4 = 20 - 80 = -60 \), indicating \( f \) is decreasing.
04

Identify relative maxima and minima

Based on the intervals of increase and decrease, we can identify relative extrema:- \( x = -\sqrt{2} \) changes from decreasing to increasing, indicating a relative minimum.- \( x = \sqrt{2} \) changes from increasing to decreasing, indicating a relative maximum.
05

Sketch the rough graph

Based on the critical points, extremas, and the nature of the function, sketch a rough graph of \( f(x) = 5 + 20x - x^5 \). - The graph starts decreasing from the left towards \( x = -\sqrt{2} \), reaches a minimum, then increases until \( x = \sqrt{2} \), where it reaches a maximum, and finally decreases again.

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Key Concepts

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

First Derivative
The first derivative of a function is a powerful tool in calculus that helps us understand how the function behaves at any given point. It essentially tells us the slope of the function at each point. For the function given, we need to determine how it increases or decreases over its domain. To start, we compute the first derivative of the function \( f(x) = 5 + 20x - x^5 \) using the power rule. This results in the derivative \[ f'(x) = 20 - 5x^4. \]
This derivative expression will help us find critical points by setting it to zero and solving for \( x \). These points, where the first derivative equals zero, usually indicate where the function changes direction either from increasing to decreasing or vice versa.
In our particular function, setting \( 20 - 5x^4 = 0 \) allows us to solve and find the critical points, which are the values of \( x \) such that \( x = \sqrt{2} \) and \( x = -\sqrt{2} \). These points are crucial as they divide the function into intervals where behavior changes. We will explore these intervals in the next section.
Increasing and Decreasing Intervals
Determining where a function is increasing or decreasing is straightforward once we have the first derivative. When\( f'(x) > 0 \), the function is increasing, and when \( f'(x) < 0 \), it's decreasing.
To identify these intervals for \( f(x) = 5 + 20x - x^5 \), we examine the sign of \( f'(x) = 20 - 5x^4 \) in the critical point-divided regions: \((-fty, -\sqrt{2})\), \((-\sqrt{2}, \sqrt{2})\), and \((\sqrt{2}, fty)\).
  • For \(x = -2\), a test point in \((-fty, -\sqrt{2})\), \( f'(-2) = -60 \), showing a decreasing trend.
  • For \(x = 0\), in \((-\sqrt{2}, \sqrt{2})\), \( f'(0) = 20 \), indicating an increasing trend.
  • For \(x = 2\), in \((\sqrt{2}, fty)\), \( f'(2) = -60 \), showing another decrease.
This means the function decreases to \( x = -\sqrt{2} \), then increases up to \( x = \sqrt{2} \), and finally decreases again beyond this point. Identifying these sequences helps us predict how the function behaves graphically and aids in comprehending changes in trends.
Relative Maxima and Minima
Relative maxima and minima refer to points on a function where it reaches a peak (maximum) or a trough (minimum) in their neighborhoods. These are discovered by analyzing where the function shifts from increasing to decreasing and vice versa.
Based on the function \( f(x) = 5 + 20x - x^5 \) and the critical points \( x = \sqrt{2} \) and \( x = -\sqrt{2} \), we gain insight:
  • At \( x = -\sqrt{2} \), the function transitions from decreasing to increasing, indicating a **relative minimum**.
  • At \( x = \sqrt{2} \), the function switches from increasing to decreasing, signaling a **relative maximum**.
Such points are pivotal because they offer locations of interest on the function curve where significant changes occur.
Thus, these relative extrema shape our understanding of the overall behavior of the function, and paired with the intervals of increase and decrease, they provide a comprehensive depiction of the graph's form. Understanding these concepts allows for skillful problem-solving and analysis of real-world scenarios modeled by functions.

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