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Chapter 4: Problem 48

Use a graphing utility to generate the graphs of \(f^{\prime}\) and \(f^{\prime \prime}\) over the stated interval; then use those graphs to estimate the \(x\)-coordinates of the inflection points of \(f,\) the intervals on which \(f\) is concave up or down, and the intervals on which \(f\) is increasing or decreasing. Check your estimates by graphing \(f\). \(f(x)=\frac{1}{1+x^{2}}, \quad-5 \leq x \leq 5\)

Short Answer

Expert verified
Inflection points at approximately \( x = \pm \sqrt{\frac{1}{3}} \), increasing on \( (-\infty, 0) \), decreasing on \( (0, \infty) \), concave up on \( ( -\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} ) \), concave down elsewhere.

Step by step solution

01

Derive the First Derivative

To find the inflection points and concavity, we must first compute the first derivative of the function. The given function is \[ f(x) = \frac{1}{1+x^2} \]. We will use the quotient rule to compute the derivative:\[ f'(x) = -\frac{2x}{(1+x^2)^2} \].
02

Derive the Second Derivative

Now we need the second derivative for concavity analysis and to find potential inflection points. Taking the derivative of \[ f'(x) = -\frac{2x}{(1+x^2)^2} \] using the quotient rule gives:\[ f''(x) = \frac{2(1+x^2)^2 - 8x^2(1+x^2)}{(1+x^2)^4} \]. Simplifying this, we get:\[ f''(x) = \frac{2-6x^2}{(1+x^2)^3} \].
03

Use Graphing Utility for Derivatives

Use a graphing tool to plot both \[ f'(x) = -\frac{2x}{(1+x^2)^2} \] and \[ f''(x) = \frac{2-6x^2}{(1+x^2)^3} \] over the interval \( -5 \leq x \leq 5 \). Observe where \( f'(x) \) is zero or changes sign for increasing/decreasing intervals, and where \( f''(x) \) changes sign for concavity and inflection points.
04

Analyze the First Derivative

From the graph of \( f'(x) \), observe where it crosses the x-axis. These points are the critical points. Determine where \( f'(x) \) is above the axis (positive) indicating that \( f \) is increasing, and where it is below the axis (negative) indicating that \( f \) is decreasing.
05

Analyze the Second Derivative

Examine the graph of \( f''(x) \) to identify where it crosses the x-axis. These points are potential inflection points. Wherever \( f''(x) \) is positive, \( f \) is concave up. Wherever \( f''(x) \) is negative, \( f \) is concave down.
06

Validate by Graphing the Function

Finally, graph the original function \( f(x) = \frac{1}{1+x^2} \) over the interval \( -5 \leq x \leq 5 \) to verify the findings regarding inflection points, concavity, and intervals of increase and decrease.

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Key Concepts

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

Derivatives
Understanding derivatives is key to analyzing functions, especially when exploring their behavior across different intervals. The derivative of a function, denoted as \( f'(x) \), provides valuable information about the rate at which the function's value changes with respect to changes in \( x \). In simpler terms, it tells us how the function is behaving over specific sections of its graph.

  • **First Derivative:** This derivative helps identify where the function is increasing or decreasing. If \( f'(x) > 0 \), the function is increasing, meaning the graph slopes upwards. Conversely, if \( f'(x) < 0 \), the function is decreasing, indicating a downward slope. Moreover, the points where \( f'(x) = 0 \) are critical; these can be potential turning points of the function's graph.
  • **Second Derivative:** The second derivative, \( f''(x) \), is crucial for understanding the concavity of the function, which refers to whether the graph curves upwards or downwards. This helps in identifying inflection points, where the concavity changes.
Mastering derivatives gives you the tools to predict and understand the inherent tendencies of mathematical functions.
Concavity
Concavity describes the direction of the curve of a graph. By analyzing the second derivative, \( f''(x) \), we can understand whether a function's graph is concave up or down. This is important for visualizing how the function behaves over certain intervals.

  • **Concave Up:** If \( f''(x) > 0 \), the graph of the function is concave up. Think of this as a bowl facing upwards, where the curve opens up like a smile. It suggests that the rate of increase (or decrease) itself is increasing.
  • **Concave Down:** Conversely, if \( f''(x) < 0 \), the graph is concave down, much like an upside-down bowl or a frown. Here, the rate of increase is decreasing, meaning the graph bends downwards.
The notion of concavity allows mathematicians and students alike to predict how steep or shallow a graph will be at different points, which can be vital in various real-world applications.
Inflection Points
Inflection points are the particular points on a graph where the concavity changes. These points can be seen as the transition markers, where the graph changes from being concave up to concave down or vice versa. Identifying these points is crucial to understanding the overall shape and behavior of a function.

  • **Finding Inflection Points:** To find potential inflection points, set \( f''(x) = 0 \) and solve for \( x \). These points indicate where to inspect for a change in concavity.
  • **Verification:** It's essential to confirm that there is an actual change in concavity. This involves checking intervals around the potential inflection points to ensure the sign of \( f''(x) \) changes from positive to negative, or vice versa.
Understanding inflection points helps in designing curves and ensuring calculations accurately represent real-world phenomena or intended outcomes in mathematical modeling.

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Most popular questions from this chapter

Let \(s(t)=5 t^{2}-22 t\) be the position function of a particle moving along a coordinate line, where \(s\) is in feet and \(t\) is in seconds. (a) Find the maximum speed of the particle during the time interval \(1 \leq t \leq 3\). (b) When, during the time interval \(1 \leq t \leq 3,\) is the particle farthest from the origin? What is its position at that instant?

(a) Show that if \(f\) is differentiable on \((-\infty,+\infty)\), and if \(y=f(x)\) and \(y=f^{\prime}(x)\) are graphed in the same \(c 0\) -ordinate system, then between any two \(x\) -intercepts of \(f\) there is at least one \(x\) -intercept of \(f^{\prime}\). (b) Give some examples that illustrate this.

Use a graphing utility to determine how many solutions the equation has, and then use Newton鈥檚 Method to approximate the solution that satisfies the stated condition. \(x-\tan x=0 ; \pi / 2

If \(f\) is a periodic function, then the locations of all absolute extrema on the interval \((-\infty,+\infty)\) can be obtained by finding the locations of the absolute extrema for one period and using the periodicity to locate the rest. Use this idea in these exercises to find the absolute maximum and minimum values of the function, and state the \(x\) -values at which they occur. \(f(x)=3 \cos \frac{x}{3}+2 \cos \frac{x}{2}\)

Prove that a particle is speeding up if the velocity and acceleration have the same sign, and slowing down if they have opposite signs. [Hint: Let \(r(t)=|v(t)|\) and find \(r^{\prime}(t)\) using the chain rule.]
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