Problem 46 - Essential Calculus

Question

Find the critical numbers of \(f\) (if any). Find the open intervals on which the function is increasing or decreasing and locate all relative extrema. Use a graphing utility to confirm your results. $$ f(x)=\frac{x^{3}}{3}-\ln x $$

Step-by-Step Solution

Answer

The critical number of the function is \(x=1\). The function is decreasing in the interval \((- \infty, 1)\) and increasing in the interval \((1, \infty)\). The function has a local minimum at \(x=1\).

Step 1: Compute Derivative of \(f\)

The first step is to compute the derivative of the function, \(f(x) = \frac{x^{3}}{3} - \ln x\). This gives the derivative as \(f'(x) = x^{2} - \frac{1}{x}\).

Step 2: Find Critical Numbers

Critical numbers are found by setting the derivative equal to zero. So, solve for \(x\) in the equation \(f'(x) = 0\) to get the critical numbers. This gives us \(x^{2} - \frac{1}{x} =0\). Multiplying by x to eliminate the fraction provides \(x^{3} -1=0\) which gives us the critical number \(x= 1\).

Step 3: Intervals of Increase and Decrease

To find where the function is increasing or decreasing, study the sign of \(f'(x)\) on the intervals \((- \infty, 1)\) and \((1, \infty)\). Just choose random numbers from these intervals and substitute these numbers into \(f'(x)\). If \(f'(x) > 0\), the function is increasing. If \(f'(x) < 0\), the function is decreasing. For \(x=0.5\), \(f'(0.5) < 0\); and for \(x=2, f'(x) > 0\). Thus the function is decreasing on \((- \infty, 1)\) and increasing on \((1, \infty)\).

Step 4: Determine Relative Extrema

As the function \(f\) is decreasing before \(x=1\) and increasing after \(x=1\), this means that at \(x=1\), \(f\) has a local minimum. Therefore, the only relative extremum is the minimum at \(x=1\).

Key Concepts

Increasing and Decreasing Intervals

When studying a function, it's valuable to understand where the function increases or decreases. These intervals can provide insight into the behavior and shape of the function's graph. To find these intervals, we must first determine the function's derivative. The derivative,

• tells us the rate at which the function's value changes
• can indicate where a function is moving upwards (increasing) or downwards (decreasing)

For the function \( f(x) = \frac{x^{3}}{3} - \ln x \), the derivative is \( f'(x) = x^{2} - \frac{1}{x} \). To find the increasing or decreasing intervals:
- Identify the critical numbers by solving \( f'(x) = 0 \).
- Next, choose test points from each interval created by these critical numbers.
Substitute these test points into \( f'(x) \):

• If \( f'(x) > 0 \), the function is increasing in that interval.
• If \( f'(x) < 0 \), the function is decreasing in that interval.

For example, by testing values like \( 0.5 \) and \( 2 \), we found that our function is decreasing on \((- \infty, 1)\) and increasing on \((1, \infty)\).

Relative Extrema

Relative extrema in a function are the points where the function reaches a local maximum or minimum value. These points indicate where a graph turns upward or downward and are essential for understanding a function's shape. To locate relative extrema:

• Identify critical numbers where \( f'(x) = 0 \) or \( f'(x) \) is undefined.
• Determine the function's behavior around these points using test values.

For example, in the function \( f(x) = \frac{x^{3}}{3} - \ln x \), we found a critical number at \( x = 1 \).
By analyzing the intervals around this critical number (\(-\infty, 1\) and \(1, \infty\)), we determined that the function changes from decreasing to increasing at \( x = 1 \). This signifies the presence of a local minimum. Hence, the only relative extremum for our function is a local minimum at \( x = 1 \).

Derivatives

Derivatives are fundamental concepts in calculus that describe how a function changes. They give insights into:

• the slope of the function at any point
• the rate of change of the function
• whether the function is increasing or decreasing

In our example, the derivative of \( f(x) = \frac{x^{3}}{3} - \ln x \) is \( f'(x) = x^{2} - \frac{1}{x} \).
This equation tells us:

• The function's slope, indicating how steep or flat the curve is
• Critical points where \( f'(x) = 0 \) help locate potential maxima or minima

To compute a derivative, apply rules of differentiation to the original function. Identify ingredients like power rule for polynomial terms and chain rule when necessary. Understanding derivatives allows us to paint a comprehensive picture of a function's geometry and its dynamics across the graph.