Problem 10 - Thomas Calculus in SI Units

Question

Answer the following questions about the functions whose derivatives are given in Exercises \(1-14:\) a. What are the critical points of \(f ?\) b. On what open intervals is \(f\) increasing or decreasing? c. At what points, if any, does \(f\) assume local maximum and minimum values? $$f^{\prime}(x)=3-\frac{6}{\sqrt{x}}, \quad x \neq 0$$

Step-by-Step Solution

Answer

Critical point at \(x=4\), \(f\) decreases on \((0, 4)\) and increases on \((4, \infty)\), with a local minimum at \(x=4\).

Step 1: Set the Derivative to Zero

To find the critical points, set the derivative equal to zero: \(f'(x) = 3-\frac{6}{\sqrt{x}} = 0\). Solve for \(x\).

Step 2: Solve for Critical Points

Rearrange the equation from Step 1 to get \(3 = \frac{6}{\sqrt{x}}\). Multiply both sides by \(\sqrt{x}\) to isolate it: \(3\sqrt{x} = 6\). Divide by 3: \(\sqrt{x} = 2\). Square both sides: \(x = 4\). The critical point is \(x=4\).

Step 3: Find Intervals for Increasing/Decreasing

To determine where \(f\) is increasing or decreasing, test intervals around the critical point. Choose test points from the intervals \((0, 4)\) and \((4, \infty)\) to plug into \(f'(x)\).

Step 4: Test Sign of Derivative

Choose test points, such as \(x = 1\) for \((0, 4)\) and \(x = 9\) for \((4, \infty)\).- For \(x = 1\): \( f'(1) = 3 - \frac{6}{\sqrt{1}} = 3 - 6 = -3\). So \(f\) is decreasing on \((0, 4)\).- For \(x = 9\): \( f'(9) = 3 - \frac{6}{\sqrt{9}} = 3 - 2 = 1\). So \(f\) is increasing on \((4, \infty)\).

Step 5: Determine Local Extrema

Since \(f\) changes from decreasing to increasing at \(x=4\), \(f\) has a local minimum at \(x=4\).

Key Concepts

Increasing and Decreasing Intervals

Understanding whether a function is increasing or decreasing on certain intervals is a key part of analyzing the function's behavior. This can be determined by looking at the sign of the derivative. If the derivative, denoted as \(f'(x)\), is positive on an interval, the function is increasing on that interval. Conversely, if \(f'(x)\) is negative, the function is decreasing.

In our example, the derivative is given by \(f'(x) = 3 - \frac{6}{\sqrt{x}}\). To find where the function changes behavior, we examine intervals around the critical point found in the steps. We explored two intervals: \((0, 4)\) and \((4, \infty)\). By selecting a test point in each interval, we determine the sign of \(f'(x)\):

• For interval \((0, 4)\), the test point \(x = 1\) gives \(f'(1) = -3\), indicating that the function is decreasing.
• For interval \((4, \infty)\), the test point \(x = 9\) gives \(f'(9) = 1\), indicating that the function is increasing.

Therefore, the function decreases on \((0, 4)\) and increases on \((4, \infty)\).

Local Maximum and Minimum

The concept of local maxima and minima involves identifying points where a function reaches a high or low value within a neighboring interval. A function has a local maximum if it changes from increasing to decreasing, and a local minimum if it changes from decreasing to increasing.

In our exercise, we found that \(f\) has a critical point at \(x = 4\) using the derivative \(f'(x) = 3 - \frac{6}{\sqrt{x}}\). Testing the sign of \(f'(x)\) on either side of this critical point reveals:

• On \((0, 4)\), \(f'(x)\) is negative, indicating the function is decreasing.
• On \((4, \infty)\), \(f'(x)\) is positive, indicating the function is increasing.

Since the function transitions from decreasing to increasing at \(x = 4\), this point is identified as a local minimum. There is no local maximum in this particular function as there is no point that was transitioning from increasing to decreasing.

Derivative Test

The Derivative Test is a mathematical method used to identify critical points of a function and to determine whether those points are local maxima, minima, or neither. It relies on examining the derivative \(f'(x)\) of a function.

To apply the derivative test, follow these steps:

• First, find the derivative \(f'(x)\) of the function.
• Set \(f'(x) = 0\) and solve for \(x\) to find the critical points. These are points where the function might change behavior.
• Determine the sign of \(f'(x)\) on intervals around each critical point.

In our case, we calculated \(f'(x)\) as \(3 - \frac{6}{\sqrt{x}}\) and found the critical point at \(x = 4\). Testing the intervals showed that before this point, the function was decreasing, and after this point, it was increasing. Therefore, by the Derivative Test, \(x = 4\) is a local minimum. This test is vital as it provides a structured approach to finding local extremum points.