Problem 36 - Thomas Calculus in SI Units

Question

a. Find the open intervals on which the function is increasing and decreasing. b. Identify the function's local and absolute extreme values, if any, saying where they occur. $$f(x)=\frac{x^{3}}{3 x^{2}+1}$$

Step-by-Step Solution

Answer

The function decreases on \((-\infty, 0)\) and increases on \((0, \infty)\), with a local minimum at \(x = 0\).

Step 1: Find the Derivative

To determine where the function is increasing or decreasing, we first find the derivative of the function. Given the function \( f(x) = \frac{x^{3}}{3x^{2}+1} \), we apply the quotient rule. The formula for the derivative of a quotient \( \frac{u}{v} \) is \( \frac{u'v - uv'}{v^2} \). Here, \( u = x^3 \) and \( v = 3x^2 + 1 \). Thus: \[ f'(x) = \frac{(3x^2)(3x^2 + 1) - (x^3)(6x)}{(3x^2 + 1)^2} \] Simplify this expression step by step.

Step 2: Simplify the Derivative

Continue simplifying: \[ f'(x) = \frac{9x^4 + 3x^2 - 6x^4}{(3x^2 + 1)^2} \]Combine like terms: \[ f'(x) = \frac{3x^4 + 3x^2}{(3x^2 + 1)^2} \]Factor out the common terms: \[ f'(x) = \frac{3x^2(x^2 + 1)}{(3x^2 + 1)^2} \]

Step 3: Find Critical Points

To find critical points, set the derivative equal to zero and solve for \( x \):\[ 3x^2(x^2 + 1) = 0 \]This gives us \( x = 0 \). Also, consider where the derivative does not exist, but since the denominator \( (3x^2 + 1)^2 \) is never zero, it does not contribute additional critical points.

Step 4: Establish Intervals

Now determine the sign of \( f'(x) \) in different intervals divided by the critical point. Check intervals \( (-\infty, 0) \) and \( (0, \infty) \):- For \( x \in (-\infty, 0) \): Choose \( x = -1 \), \( f'(-1) = \frac{-3}{4} < 0 \), so \( f(x) \) is decreasing.- For \( x \in (0, \infty) \): Choose \( x = 1 \), \( f'(1) = \frac{3}{4} > 0 \), so \( f(x) \) is increasing.

Step 5: Identify Extremes

Since the function changes from decreasing to increasing at \( x = 0 \), there is a local minimum at \( x = 0 \). Calculate \( f(0) \):\[ f(0) = \frac{0}{3(0)^2 + 1} = 0 \]Therefore, \( f(x) \) has a local minimum value of 0 at \( x = 0 \). No absolute maximum or minimum is present without further bounds on \( x \).

Key Concepts

Critical Points

Critical points are vital in analyzing the behavior of a function. A critical point is a value of \( x \) where the derivative of the function equals zero or does not exist. This is important because critical points often highlight where a function's behavior changes, such as shifting from increasing to decreasing or vice versa.
In our given function \( f(x) = \frac{x^{3}}{3x^{2}+1} \), we found the derivative \( f'(x) \) using the quotient rule:

• \( f'(x) = \frac{3x^2(x^2 + 1)}{(3x^2 + 1)^2} \)

To locate the critical points, we set \( f'(x) = 0 \). This simplifies to \( 3x^2(x^2 + 1) = 0 \), which gives the critical point \( x = 0 \). Since the denominator \( (3x^2 + 1)^2 \) is never zero, no additional critical points arise from within the fraction.

Extreme Values

Extreme values, either local or absolute, indicate where a function reaches its highest or lowest points within a certain range. Local extrema are points where the function value is a local maximum or minimum, while absolute extrema are the highest and lowest points over the function’s entire domain.For the function \( f(x) = \frac{x^{3}}{3x^{2}+1} \), we discovered a local minimum. Since the derivative changes sign from negative to positive at \( x = 0 \), it indicates a local minimum at this point. By substituting into the original function:

• \( f(0) = 0 \)

This confirms the local minimum value of 0 at \( x = 0 \). Because the function's behavior at infinity or a broader domain isn’t specified, no absolute extremes are identified.

Increasing and Decreasing Intervals

Understanding where a function increases or decreases is crucial for understanding its overall behavior. A function is increasing on an interval if its derivative is positive over the interval and decreasing if its derivative is negative.After finding \( f'(x) = \frac{3x^2(x^2 + 1)}{(3x^2 + 1)^2} \), we checked the sign of the derivative in the intervals divided by the critical point:

• For \( x \in (-\infty, 0) \), the derivative is negative, as seen by choosing \( x = -1 \) (\( f'(-1) = \frac{-3}{4} \)). Thus, the function is decreasing here.
• For \( x \in (0, \infty) \), with \( x = 1 \), the derivative is positive (\( f'(1) = \frac{3}{4} \)), indicating the function is increasing.

These findings highlight how the function changes at \( x = 0 \), from decreasing to increasing.

Local Minimum

A local minimum is a point where a function has a lower value than at all other points in its immediate vicinity. It doesn't need to be the lowest value of the whole function, just the lowest in its neighborhood.In our analysis of \( f(x) = \frac{x^{3}}{3x^{2}+1} \), we identified a local minimum at \( x = 0 \). This conclusion is derived from the derivative test:

• The function transitions from decreasing (negative derivative) to increasing (positive derivative) at \( x = 0 \).
• Evaluating \( f(0) \) confirms a local minimum value of 0.

This local minimum doesn't translate to an absolute minimum over all possible values, but it is a critical indicator of the function's topology around \( x = 0 \). Such points are essential in optimization problems and understanding complicated graphs.