Problem 26 - Calculus Early Transcendentals

Question

Find: (a) the intervals on which f is increasing, (b) the intervals on which f is decreasing, (c) the open intervals on which f is concave up, (d) the open intervals on which f is concave down, and (e) the x-coordinates of all inflection points. \(f(x)=x^{2 / 3}-x\)

Step-by-Step Solution

Answer

(a) None. (b) \((-\infty, \infty)\). (c) None. (d) \((0, \infty)\). (e) None.

Step 1: Find the Derivative

First, let's find the first derivative of the function \(f(x) = x^{2/3} - x\). The derivative is found using the power rule for differentiation. - The derivative of \(x^{2/3}\) is \(\frac{2}{3}x^{-1/3}\).- The derivative of \(x\) is 1.Thus, the first derivative is:\[ f'(x) = \frac{2}{3}x^{-1/3} - 1 \]

Step 2: Find Critical Points

To find the critical points, set the first derivative equal to zero:\[ \frac{2}{3}x^{-1/3} - 1 = 0 \]Solving for \(x\):\[ \frac{2}{3}x^{-1/3} = 1 \]\[ x^{-1/3} = \frac{3}{2} \]Taking both sides to the power of -3 to solve for \(x\):\[ x = \left(\frac{3}{2}\right)^{-3} = \frac{8}{27} \]

Step 3: Analyze Critical Points and Increasing/Decreasing Intervals

Use test points to determine where \(f'(x) > 0\) or \(f'(x) < 0\).- Choose a test point to the left of \(x = \frac{8}{27}\), say \(x = 0\): \(f'(0) = \frac{2}{3}\times 0^{-1/3} - 1 = -1\) (negative).- Choose a test point to the right of \(x = \frac{8}{27}\), say \(x = 1\): \(f'(1) = \frac{2}{3}(1)^{-1/3} - 1 = \frac{2}{3} - 1 = -\frac{1}{3}\) (negative).Since \(f'(x)\) is negative on both intervals, there are no intervals where \(f(x)\) is increasing. It is decreasing on the interval \((-\infty, \infty)\).

Step 4: Find the Second Derivative

To analyze concavity, find the second derivative:We previously found \(f'(x) = \frac{2}{3}x^{-1/3} - 1\). Now, differentiate again:\[ f''(x) = \frac{d}{dx}\left(\frac{2}{3}x^{-1/3}\right) = \frac{2}{3}\cdot (-\frac{1}{3}x^{-4/3}) = -\frac{2}{9}x^{-4/3} \]

Step 5: Determine Concave Up and Concave Down Intervals

Check where \(f''(x) > 0\) and \(f''(x) < 0\):- \(f''(x)\) is negative for \(x > 0\) since \(-\frac{2}{9}x^{-4/3} < 0\).- \(f''(x)\) is undefined for \(x \leq 0\).Therefore, \(f(x)\) is concave down on \((0, \infty)\) and it does not exhibit concave up behavior anywhere.

Step 6: Find Inflection Points

An inflection point occurs where the concavity changes. However, since \(f''(x)\) does not change sign and is undefined for \(x \leq 0\), there are no inflection points.

Key Concepts

Derivatives

Derivatives are a fundamental concept in calculus that help us understand how functions change. The derivative of a function at a given point measures its instantaneous rate of change. To find the derivative of a function like \[ f(x) = x^{2/3} - x \]use differentiation rules to simplify it. This step requires applying the power rule, useful for handling exponents such as fractions and whole numbers.

• The derivative of \( x^{2/3} \) is \( \frac{2}{3}x^{-1/3} \).
• The derivative of \( -x \) is \(-1\).

Combining these, we get: \[ f'(x) = \frac{2}{3}x^{-1/3} - 1 \]This expression allows us to explore where the function is increasing or decreasing. It is a cornerstone for finding critical points and analyzing the function further.

Concavity

Concavity refers to the "curviness" or bending nature of a graph. A graph, or a function, can either be concave up, resembling a cup that holds water, or concave down, like an arch that would let water slide off. To determine concavity, look at the second derivative of the function.If the second derivative is positive over an interval, the function is concave up on that interval. Conversely, if it is negative, the function is concave down. For \[ f(x) = x^{2/3} - x \]we calculate:\[ f''(x) = -\frac{2}{9}x^{-4/3} \]In this case, \( f''(x) \) is negative for \( x > 0 \), indicating the function is concave down in the interval \((0,\infty)\). It is crucial to know that concavity tells us about the overall shape and bending direction of a curve but doesn't directly provide where it starts or ends.

Critical Points

Critical points are where a function's derivative equals zero or is undefined. These points are essential because they are where potential changes in the function's behavior, such as turning points or peaks and valleys, occur. Finding the critical points for the function \[ f(x) = x^{2/3} - x \]involves setting the first derivative equal to zero:\[ \frac{2}{3}x^{-1/3} - 1 = 0 \]Solving this gives:\[ x = \left(\frac{3}{2}\right)^{-3} = \frac{8}{27} \]This value is a critical point. Analyze regions around this point to determine if the function is increasing or decreasing. For this example, the derivative does not change sign around \( x = \frac{8}{27} \), meaning that the function decreases on its entire domain, \(( -\infty, \infty) \). These points are crucial in sketching graphs and understanding function behavior.

Inflection Points

Inflection points occur where a function's concavity changes direction, from concave up to concave down or vice versa. They indicate critical changes in a curve's bending and are essential for understanding the overall shape of a graph. For the function \[ f(x) = x^{2/3} - x \],we check where the second derivative \[ f''(x) = -\frac{2}{9}x^{-4/3} \]changes sign.However, in this case, \( f''(x) \) remains negative for all \( x > 0 \) and is undefined for \( x \leq 0 \). Therefore, no points satisfy the criteria for an inflection point since the concavity does not change. Identifying inflection points is crucial for drawing accurate graphs and understanding the nature of a function's concavity across its domain.