Problem 24 - Thomas Calculus

Question

In Exercises \(9-28 :\) a. Find the intervals on which the function is increasing and decreasing. b. Then identify the function's local extreme values, if any, saying where they are taken on. c. Which, if any, of the extreme values are absolute? d. Support your findings with a graphing calculator or computer grapher. $$ f(x)=\frac{x^{3}}{3 x^{2}+1} $$

Step-by-Step Solution

Answer

The function is increasing on \((-1,1)\), decreasing on \((-\infty,-1)\) and \((1,\infty)\). Local min at \(x=-1\), local and absolute max at \(x=1\).

Step 1: Find Derivative of the Function

To determine where the function is increasing or decreasing, first find the first derivative of the function, which is necessary to identify the critical points and intervals. The given function is \( f(x) = \frac{x^3}{3x^2+1} \). Apply the quotient rule to find \( f'(x) \): \[f'(x) = \frac{(3x^2)(3x^2+1) - x^3(6x)}{(3x^2+1)^2} = \frac{3x^4 + 3x^2 - 6x^4}{(3x^2+1)^2}\]Simplify the derivative:\[f'(x) = \frac{-3x^4 + 3x^2}{(3x^2+1)^2} = \frac{3x^2(1-x^2)}{(3x^2+1)^2}\]

Step 2: Critical Points

Find the critical points by setting \( f'(x) = 0 \) and solving for \( x \):\[3x^2(1-x^2) = 0\]This yields \( x = 0 \) and \( x = \pm 1 \) as critical points. The function is only defined when the denominator is not zero, but \( 3x^2 + 1 eq 0 \) for all real \( x \). Hence, the critical points are valid.

Step 3: Test Intervals for Increasing/Decreasing Behavior

Test intervals around the critical points \( x = 0 \) and \( x = \pm 1 \):- For \( x < -1 \), pick \( x = -2 \): \( f'(-2) < 0 \) so the function decreases.- For \( -1 < x < 0 \), pick \( x = -0.5 \): \( f'(-0.5) > 0 \) so the function increases.- For \( 0 < x < 1 \), pick \( x = 0.5 \): \( f'(0.5) > 0 \) so the function increases.- For \( x > 1 \), pick \( x = 2 \): \( f'(2) < 0 \) so the function decreases.

Step 4: Determine Local Extreme Values

Use the first derivative test to determine local minima and maxima:- At \( x = -1 \), the function changes from decreasing to increasing, hence a local minimum.- At \( x = 0 \), the function continues increasing, so no local extremum here.- At \( x = 1 \), the function changes from increasing to decreasing, hence a local maximum.

Step 5: Identify Absolute Extrema

Consider the behavior of \( f(x) \) as \( x \to \pm \infty \). The function tends towards zero. Since the function is defined for all real \( x \), and the local maximum at \( x = 1 \) was found, check its value:\[f(1) = \frac{1^3}{3(1)^2+1} = \frac{1}{4}\]Thus, \( x = 1 \) is a local maximum of \( \frac{1}{4} \), and as the function's limits don't exceed \( \frac{1}{4} \), this maximum is also absolute.

Step 6: Graphing the Function

Use a graphing calculator to plot \( f(x) = \frac{x^3}{3x^2+1} \) and verify:- The curve should descend for \( x < -1 \), ascend in \(-1 < x < 0 \), continue ascending\(0 < x < 1 \), and descend again for \( x > 1 \).- The graph should confirm a local minimum at \( x = -1 \) and an absolute maximum at \( x = 1 \).

Key Concepts

Derivative

The derivative is a fundamental concept in calculus that tells us how a function changes at any given point. For a function \( f(x) = \frac{x^3}{3x^2+1} \), the derivative, denoted as \( f'(x) \), is crucial to understand the behavior of the function. It essentially provides the rate of change or the slope of the tangent line at any point on the function's curve.
In this exercise, to find \( f'(x) \), we use the quotient rule, which applies when you have a ratio of two functions \( u(x) \) and \( v(x) \). The rule is stated as: \[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2} \] For our function, \( u(x) = x^3 \) and \( v(x) = 3x^2 + 1 \). Applying this rule yields the derivative: \( f'(x) = \frac{3x^2 (3x^2 + 1) - x^3(6x)}{(3x^2 + 1)^2} \).
Understanding and simplifying this derivative gives us insights into the function's behavior, especially when checking for increasing or decreasing intervals.

Critical Points

Critical points are specific values of \( x \) where the derivative \( f'(x) \) is zero or undefined. These points are essential as they indicate potential local maxima or minima, which are pivotal when analyzing the function's behavior.
In this example, we set the simplified form of the derivative \( f'(x) = \frac{3x^2(1-x^2)}{(3x^2+1)^2} \) to zero to find critical points: \[ 3x^2(1-x^2) = 0 \] Solving this equation, we find the critical points at \( x = 0 \), \( x = 1 \), and \( x = -1 \).
At these critical points, the function does something interesting - the slope is zero, indicating horizontal tangents, suggesting potential peaks or valleys or other changes in direction.

Increasing and Decreasing Functions

To determine where the function is increasing or decreasing, we use the first derivative test. By analyzing the sign of \( f'(x) \) in intervals around the critical points, we can conclude where the function rises or falls.
- Between critical points \( x = -\infty \) to \( x = -1 \), the derivative is negative, meaning the function decreases.
- From \( x = -1 \) to \( x = 0 \), the derivative is positive, so the function increases.
- Between \( x = 0 \) to \( x = 1 \), the derivative remains positive, continuing the increase.
- Beyond \( x = 1 \) to \( x = \infty \), the derivative reverts to being negative, indicating a decrease again.
This analysis tells us the ups and downs of the function without needing to plot it visually, effectively summarizing the slopes along the curve.

Local Extreme Values

Local extreme values, which include local minima and maxima, are the points on the graph where the function switches its behavior from increasing to decreasing, or vice versa.
Using the first derivative test again, we find that:

• At \( x = -1 \), the function shifts from decreasing to increasing, identifying a local minimum.
• At \( x = 1 \), it changes from increasing to decreasing, marking a local maximum.

Beyond just finding these, evaluating the function's end behavior and extreme values helps in identifying any absolute extremes. Here, it was determined that the local maximum at \( x = 1 \) is also an absolute maximum with the value \( \frac{1}{4} \). In contrast, as the function approaches infinity, it stabilizes and approaches zero but never surpasses the observed local maximum, affirming its absoluteness.