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Chapter 4: Problem 10

Answer the following questions about the functions whose derivatives are given. 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 or minimum values? $$f^{\prime}(x)=3-\frac{6}{\sqrt{x}}, \quad x \neq 0$$

Short Answer

Expert verified
a. Critical point: \( x = 4 \). b. \( f(x) \) is decreasing on \( (0, 4) \) and increasing on \( (4, \infty) \). c. \( f(x) \) has a local minimum at \( x = 4 \).

Step by step solution

01

Find the Critical Points

To find the critical points of the function, we set the derivative equal to zero. Thus, we solve the equation \( 3 - \frac{6}{\sqrt{x}} = 0 \). This implies \( \frac{6}{\sqrt{x}} = 3 \). Solving for \( x \), square both sides: \( \left(\frac{6}{3}\right)^2 = \sqrt{x}^2 \). Therefore, \( 2 = \sqrt{x} \), and squaring both sides gives \( x = 4 \). Since the derivative does not exist for \( x = 0 \), we exclude it as a critical point. The critical point is \( x = 4 \).
02

Determine Intervals of Increase or Decrease

To determine where the function is increasing or decreasing, we consider the sign of the derivative on intervals divided by the critical point. We consider the intervals \((0, 4)\) and \((4, \infty)\). - For \( x \in (0, 4) \), choose \( x = 1 \). Then \( f'(1) = 3 - \frac{6}{\sqrt{1}} = 3 - 6 = -3 \). Hence, \( f(x) \) is decreasing on \( (0, 4) \). - For \( x \in (4, \infty) \), choose \( x = 9 \). Then \( f'(9) = 3 - \frac{6}{\sqrt{9}} = 3 - 2 = 1 \). Hence, \( f(x) \) is increasing on \( (4, \infty) \).
03

Identify Local Maximum or Minimum

To identify local maxima or minima, we analyze the sign change of the derivative at the critical point \( x = 4 \). - Since \( f'(x) \) changes from negative to positive at \( x = 4 \), the function switches from decreasing to increasing, indicating a local minimum at \( x = 4 \).

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Key Concepts

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

Interval Notation
Interval notation is a shorthand to express the domain of functions. It tells you exactly where a function is defined or specific behavior occurs, such as increasing or decreasing.Critical points divide the number line into intervals that discuss the behavior of a function. Let's look at how this notation is applied in the exercise:
  • The critical point found was at \( x = 4 \).
  • We analyze two intervals: \((0, 4)\) and \((4, \infty)\).
  • Notice that the interval \((0, 4)\) does not include 0 or 4 as actual values for \( x \), because it is an open interval, using round brackets.
  • For \[(4, \infty)\], this indicates the function behavior towards infinity, again excluding 4 itself.
The use of open intervals, such as \((0, 4)\), signifies that endpoints are not part of the interval, which is crucial for accurately capturing real-world scenarios and behavior at critical points.
Function Analysis
Function analysis is a method used to determine how and where a function is increasing or decreasing. This involves looking at the sign of the function's derivative, which can tell us how the function behaves over specific intervals.
In this exercise, the derivative of our function is given as \(f'(x) = 3 - \frac{6}{\sqrt{x}}\). Here鈥檚 what we did:
  • First, we solved \(f'(x) = 0\) to find the critical point where the function might change direction.
  • Then, we tested points in the intervals determined by these critical points to see whether the derivative is positive or negative.
  • If \(f'(x) > 0\) in an interval, the function is increasing over that interval. Conversely, if \(f'(x) < 0\), it is decreasing.
In our problem:
- \((0, 4)\) had a negative derivative, meaning the function is decreasing.- \((4, \infty)\) had a positive derivative, meaning the function is increasing.Function analysis through its derivative helps us deeply understand how the graph of a function behaves over various segments.
Local Extrema
Local extrema refer to the points in a function where it reaches a local maximum or minimum. These points indicate where the function transitions from increasing to decreasing or vice versa. We explore how to identify these points:
  • They are generally identified at critical points, where the derivative is zero or undefined.
  • In this exercise, the critical point \(x = 4\) is examined for any local extremum.
  • A local minimum or maximum occurs based on the sign change of the derivative around the critical point.
The derivative at \(x = 4\) changed from negative to positive, indicating that after being on a downward slope, the function starts to rise again. Hence, a local minimum exists at \(x = 4\). Recognizing local extrema helps in sketching the graph of functions and understanding their peaks and troughs on specified intervals.

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Most popular questions from this chapter

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