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Chapter 4: Problem 25

25-36 $$ \begin{array}{l}{\text { (a) Find the intervals of increase or decrease. }} \\\ {\text { (b) Find the local maximum and minimum values. }} \\ {\text { (c) Find the intervals of concavity and the inflection points. }} \\ {\text { (d) Use the information from parts (a)- (c) to sketch the graph. }} \\ {\text { Check your work with a graphing device if you have one. }}\end{array} $$ $$f(x)=2 x^{3}-3 x^{2}-12 x$$

Short Answer

Expert verified
Function increases on \((-\infty, -1)\) and \((2, \infty)\), decreases on \((-1, 2)\). Max at \(x=-1\), min at \(x=2\). Inflection point at \(x=0.5\).

Step by step solution

01

Find the first derivative

To determine the intervals of increase or decrease, we need to find the first derivative of the function \( f(x) = 2x^3 - 3x^2 - 12x \). The first derivative is given by differentiating each term:\[ f'(x) = \frac{d}{dx}(2x^3) - \frac{d}{dx}(3x^2) - \frac{d}{dx}(12x) \]\[ f'(x) = 6x^2 - 6x - 12 \]
02

Find critical points

Critical points occur where the first derivative is zero or undefined. Set \( f'(x) \) equal to zero and solve for \( x \):\[ 6x^2 - 6x - 12 = 0 \]Divide the whole equation by 6 to simplify:\[ x^2 - x - 2 = 0 \]Factor the quadratic:\[ (x-2)(x+1) = 0 \]The solutions are \( x = 2 \) and \( x = -1 \). These are the critical points.
03

Determine intervals of increase/decrease

Use the critical points to test the intervals on the number line: \((-\infty, -1)\), \((-1, 2)\), \((2, \infty)\). Test a point in each interval in the first derivative to determine if the function is increasing or decreasing.- For \( x = -2 \), \( f'(-2) = 24 \), positive, so \( f(x) \) is increasing on \((-\infty, -1)\).- For \( x = 0 \), \( f'(0) = -12 \), negative, so \( f(x) \) is decreasing on \((-1, 2)\).- For \( x = 3 \), \( f'(3) = 12 \), positive, so \( f(x) \) is increasing on \((2, \infty)\).
04

Identify local maximum and minimum values

A local maximum occurs at \( x = -1 \) because the derivative changes from positive to negative. Calculate \( f(-1) \).\[ f(-1) = 2(-1)^3 - 3(-1)^2 - 12(-1) = 2 + 3 + 12 = 13 \]A local minimum occurs at \( x = 2 \) because the derivative changes from negative to positive. Calculate \( f(2) \).\[ f(2) = 2(2)^3 - 3(2)^2 - 12(2) = 16 - 12 - 24 = -20 \]
05

Find the second derivative

Determine the concavity by finding the second derivative of \( f(x) \):\[ f''(x) = \frac{d}{dx}(6x^2 - 6x - 12) \]\[ f''(x) = 12x - 6 \]
06

Determine intervals of concavity and inflection points

Find where the second derivative equals zero to locate inflection points:\[ 12x - 6 = 0 \]\[ x = \frac{1}{2} \]Test intervals \((-\infty, \frac{1}{2})\) and \((\frac{1}{2}, \infty)\) for concavity.- For \( x = 0 \), \( f''(0) = -6 \), negative, so \( f(x) \) is concave down on \((-\infty, \frac{1}{2})\).- For \( x = 1 \), \( f''(1) = 6 \), positive, so \( f(x) \) is concave up on \((\frac{1}{2}, \infty)\).Inflection point occurs at \( x = \frac{1}{2} \).
07

Sketch the graph

Using the information from previous steps, sketch the graph:- The function increases on \((-\infty, -1)\) and \((2, \infty)\).- It decreases on \((-1, 2)\).- Local maximum at \( x = -1 \) with value 13.- Local minimum at \( x = 2 \) with value -20.- Concave down on \((-\infty, \frac{1}{2})\) and concave up on \((\frac{1}{2}, \infty)\).- Inflection point at \( x = \frac{1}{2} \), changing concavity.

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Key Concepts

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

Derivatives
The concept of derivatives is pivotal to calculus and helps determine the rate of change of a function. Consider the function \( f(x) = 2x^3 - 3x^2 - 12x \). To find its derivative, or \( f'(x) \), differentiate each term individually:
  • The derivative of \( 2x^3 \) is \( 6x^2 \).
  • The derivative of \( -3x^2 \) is \( -6x \).
  • The derivative of \( -12x \) is \( -12 \).
This results in the first derivative: \( f'(x) = 6x^2 - 6x - 12 \). Knowing the derivative is essential, as it lets us analyze the behavior of the function by finding critical points and determining intervals of increase or decrease.
Critical Points
Critical points are where the first derivative of a function equals zero or is undefined. These points are essential for identifying the locations of potential relative maxima and minima. For \( f(x) \), set the first derivative \( f'(x) = 6x^2 - 6x - 12 \) equal to zero:\[ 6x^2 - 6x - 12 = 0 \]Simplify this equation by dividing through by 6:\[ x^2 - x - 2 = 0 \]Next, factor the quadratic:\[ (x-2)(x+1) = 0 \]This results in critical points at \( x = 2 \) and \( x = -1 \). Identifying these points is the basis for further analysis, like finding intervals of increase and decrease.
Intervals of Increase and Decrease
The intervals of increase and decrease tell us how the function's slope changes over different sections of its graph. After finding critical points at \( x = -1 \) and \( x = 2 \), test intervals around these points using the first derivative. Choose test points from sections such as \((-\infty, -1)\), \((-1, 2)\), and \((2, \infty)\):
  • For \( x = -2 \) 鈫� \( f'(-2) = 24 \): Positive, so \( f(x) \) is increasing on \((-\infty, -1)\).
  • For \( x = 0 \) 鈫� \( f'(0) = -12 \): Negative, so \( f(x) \) is decreasing on \((-1, 2)\).
  • For \( x = 3 \) 鈫� \( f'(3) = 12 \): Positive, so \( f(x) \) is increasing on \((2, \infty)\).
Recognizing these intervals aids in understanding how the function behaves, revealing where it rises or falls.
Concavity and Inflection Points
Concavity informs us about how a function curves鈥攚hether it bends upwards or downwards. To find concavity, compute the second derivative \( f''(x) \). For \( f(x) \), the second derivative is:\[ f''(x) = 12x - 6 \]Solve \( f''(x) = 0 \) for potential inflection points:\[ 12x - 6 = 0 \]\[ x = \frac{1}{2} \]Now, test intervals like \((-\infty, \frac{1}{2})\) and \((\frac{1}{2}, \infty)\):
  • For \( x = 0 \) 鈫� \( f''(0) = -6 \): Negative, indicating \( f(x) \) is concave down on \((-\infty, \frac{1}{2})\).
  • For \( x = 1 \) 鈫� \( f''(1) = 6 \): Positive, indicating \( f(x) \) is concave up on \((\frac{1}{2}, \infty)\).
The inflection point at \( x = \frac{1}{2} \) is critical as it marks where the function changes its curvature, enhancing our understanding of its geometry.

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