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Chapter 5: Problem 13

Find the local maxima and minima of each of the functions. Determine whether each function has local maxima and minima and find their coordinates. For each function, find the intervals on which it is increasing and the intervals on which it is decreasing. \(y=\frac{1}{3} x^{3}+\frac{1}{2} x^{2}+2, x \in \mathbf{R}\)

Short Answer

Expert verified
Local maximum at \((-1, \frac{13}{6})\); local minimum at \((0, 2)\); increasing on \((-1, 0) \cup (0, \infty)\); decreasing on \((-\infty, -1)\).

Step by step solution

01

Find the Derivative

First, we find the derivative of the given function. The function is \( y = \frac{1}{3} x^{3} + \frac{1}{2} x^{2} + 2 \). The derivative, denoted as \( y' \) or \( f'(x) \), is obtained by differentiating each term with respect to \( x \). \[ y' = x^2 + x \]
02

Set the Derivative to Zero

To find the critical points, set the derivative \( y' = x^2 + x \) equal to zero:\[ x^2 + x = 0 \] Factor the equation to solve for \( x \): \[ x(x+1) = 0 \] This gives the critical points \( x = 0 \) and \( x = -1 \).
03

Determine the Nature of Critical Points

To find whether the critical points are local minima or maxima, perform the second derivative test. Find the second derivative of the function:\[ y'' = 2x + 1 \]Evaluate the second derivative at each critical point:- For \( x = 0 \): \( y''(0) = 2(0) + 1 = 1 \). Since \( y''(0) > 0 \), \( x = 0 \) is a point of local minima.- For \( x = -1 \): \( y''(-1) = 2(-1) + 1 = -1 \). Since \( y''(-1) < 0 \), \( x = -1 \) is a point of local maxima.
04

Determine Coordinates of Local Maxima and Minima

Now calculate the function value \( y \) at the critical points to get their coordinates:- At \( x = 0 \): \( y(0) = \frac{1}{3}(0)^3 + \frac{1}{2}(0)^2 + 2 = 2 \). So the local minimum is at \( (0, 2) \).- At \( x = -1 \): \( y(-1) = \frac{1}{3}(-1)^3 + \frac{1}{2}(-1)^2 + 2 = \frac{-1}{3} + \frac{1}{2} + 2 = \frac{13}{6} \). So the local maximum is at \( (-1, \frac{13}{6}) \).
05

Determine Increasing and Decreasing Intervals

The first derivative \( y' = x^2 + x \) tells us about the increasing and decreasing intervals:- For \( x(x + 1) > 0 \), the function is increasing.- For \( x(x + 1) < 0 \), the function is decreasing.By analyzing the sign changes:- The function decreases on \((-\infty, -1)\).- It increases on \((-1, 0) \cup (0, \infty)\).

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Key Concepts

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

Derivative
In calculus, finding the derivative of a function is like discovering the formula that tells us how the function behaves at any given point. The derivative provides the rate at which the function's value changes as its input changes. For a function given by \( y = \frac{1}{3}x^3 + \frac{1}{2}x^2 + 2 \), the derivative is calculated by differentiating the function with respect to \( x \).
  • The power rule states that for \( x^n \), the derivative is \( nx^{n-1} \).
  • Applying this to each term, we find \( y' = x^2 + x \).
This expression \( y' \) or \( f'(x) \) helps us understand how fast the function is growing or shrinking at different points.
Local Maxima and Minima
Local maxima and minima are the peaks and valleys in the graph of a function. These are specific points where the function changes direction, either from increasing to decreasing or vice versa. To identify these critical points:
  • First, find where the derivative \( y' = x^2 + x \) equals zero, i.e., solve \( x^2 + x = 0 \). This gives the values for \( x = 0 \) and \( x = -1 \).
  • The next step is to determine the nature of these points. This is done using the second derivative test.
  • Solving for the second derivative, we get \( y'' = 2x + 1 \).
  • At \( x = 0 \), \( y''(0) = 1 \), which is greater than zero, indicating a local minimum at point \( (0, 2) \).
  • At \( x = -1 \), \( y''(-1) = -1 \), which is less than zero, indicating a local maximum at point \( (-1, \frac{13}{6}) \).
Thus, local maxima and minima help us identify where a function reaches the highest or lowest point within a particular region on its graph.
Increasing and Decreasing Intervals
Understanding increasing and decreasing intervals is crucial for analyzing the behavior of a function over different sections of its domain.
  • A function is "increasing" in intervals where its derivative \( y' \) is greater than zero.
  • Conversely, it is "decreasing" where \( y' \) is less than zero.
With our function,
  • If \( x(x + 1) = x^2 + x > 0 \), then the function is increasing.
  • Whereas, if \( x(x + 1) = x^2 + x < 0 \), the function is decreasing.
By checking intervals around critical points:
  • From \(( -\infty, -1)\), the function decreases.
  • From \((-1, 0)\) and then again from \((0, \infty)\), the function increases.
These intervals offer insight into how the function's graph climbs upwards or dips downwards over its span.

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