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Chapter 5: Problem 29

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. \(f(x)=3 x^{5}-20 x^{3}+5\)

Short Answer

Expert verified
Critical values at \( x = -2, 0, 2 \); increasing on \((-\infty, -2)\) and \((2, \infty)\); decreasing on \((-2, 0)\) and \((0, 2)\); relative max at \( x = -2 \), relative min at \( x = 2 \).

Step by step solution

01

Find the Derivative

First, we need to find the derivative of the function \( f(x) = 3x^5 - 20x^3 + 5 \). The derivative, \( f'(x) \), is obtained by applying standard differentiation rules: \( f'(x) = 15x^4 - 60x^2 \).
02

Find Critical Values

Critical values occur where the derivative is zero or undefined. Set the derivative equal to zero and solve for \( x \): \( 15x^4 - 60x^2 = 0 \).\[15x^2 (x^2 - 4) = 0.\]This gives \( x^2 = 0 \) or \( x^2 = 4 \), which implies \( x = 0, x = -2, x = 2 \). These are the critical values.
03

Test Intervals for Increasing/Decreasing

With critical points at \( x = -2, 0, 2 \), determine where \( f'(x) \) is positive (increasing) or negative (decreasing). Choose test points in each interval formed by critical points: \((-\infty, -2), (-2, 0), (0, 2), (2, \infty)\). **- Test Point \( x = -3 \) in \((-\infty, -2)\):** \( f'(-3) = 15(-3)^4 - 60(-3)^2 = 1350 - 540 > 0 \) (Increasing).**- Test Point \( x = -1 \) in \((-2, 0)\):** \( f'(-1) = 15(-1)^4 - 60(-1)^2 = 15 - 60 < 0 \) (Decreasing).**- Test Point \( x = 1 \) in \((0, 2)\):** \( f'(1) = 15(1)^4 - 60(1)^2 = 15 - 60 < 0 \) (Decreasing).**- Test Point \( x = 3 \) in \((2, \infty)\):** \( f'(3) = 15(3)^4 - 60(3)^2 = 1350 - 540 > 0 \) (Increasing).
04

Identify Relative Extrema

Look at the sign changes of \( f'(x) \) at each critical point:- At \( x = -2 \), \( f'(x) \) changes from positive to negative: Relative Maximum.- At \( x = 0 \), \( f'(x) \) doesn't change sign (remains negative): No Relative Extrema.- At \( x = 2 \), \( f'(x) \) changes from negative to positive: Relative Minimum.
05

Sketch Rough Graph of \( f(x) \)

The function \( f(x) \) increases on \( (-\infty, -2) \) and \( (2, \infty) \), and decreases on \( (-2, 0) \) and \( (0, 2) \). There is a relative maximum at \( x = -2 \) and a relative minimum at \( x = 2 \). Sketch these trends accordingly, making sure to show behavior at critical points.

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Key Concepts

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

Derivatives
Derivatives are fundamental in calculus as they represent how a function changes at any given point. For the function \( f(x) = 3x^5 - 20x^3 + 5 \), finding the derivative, denoted as \( f'(x) \), involves applying standard differentiation rules such as the power rule. Differentiating each term separately, we get:
  • For \( 3x^5 \), use the power rule: the derivative is \( 15x^4 \).
  • For \( -20x^3 \), again apply the power rule: the derivative is \( -60x^2 \).
  • The constant \( 5 \) has a derivative of 0 since constants do not change.
The derivative is then \( f'(x) = 15x^4 - 60x^2 \). Derivatives can help determine critical values, intervals where the function increases or decreases, and identify relative extrema.
Increasing and Decreasing Intervals
Once we have the derivative \( f'(x) = 15x^4 - 60x^2 \), we can analyze where this is positive or negative to find intervals on which the function is increasing or decreasing. Critical points, where the derivative is zero or undefined, divide the real number line into intervals.
  • To find these points, solve \( 15x^2(x^2 - 4) = 0 \), which results in \( x = 0, -2, \) and \( 2 \).
  • These points are not just where \( f'(x) = 0 \) but also where behavior changes, marking starts or stops in increasing/decreasing trends.
By choosing test points in intervals around these critical values and determining the sign of \( f'(x) \):
  • If \( f'(x) > 0 \), the function is increasing.
  • If \( f'(x) < 0 \), the function is decreasing.
We find that \( f(x) \) increases on \((-\infty, -2)\) and \((2, \infty)\), and decreases on \((-2, 0)\) and \((0, 2)\). Understanding these intervals is crucial for sketching the correct behavior of the function.
Relative Extrema
Relative extrema are points on a graph where the function reaches a relative maximum or minimum compared to nearby points. These can often be found where the derivative changes sign.
  • A relative maximum occurs where \( f'(x) \) changes from positive to negative.
  • A relative minimum occurs where \( f'(x) \) changes from negative to positive.
Analyzing our critical points:
  • At \( x = -2 \), \( f'(x) \) transitions from positive to negative, indicating a relative maximum.
  • At \( x = 0 \), \( f'(x) \) remains negative, so there is no relative extremum.
  • At \( x = 2 \), \( f'(x) \) switches from negative to positive, identifying a relative minimum.
Recognizing relative extrema helps in understanding the function's shape and behavior, especially when preparing to sketch a graph. These points are significant because they identify local peaks and troughs, showcasing changes in the trend of the function.

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