Problem 26 - Calculus Single Variable

Question

Sketch the graph of each of the functions in Exercise \(25-40,\) exhibiting and labeling: a) all local and globa extrema; b) inflection points; c) intervals on which the func tion is increasing or decreasing; d) intervals on which the function is concave up or concave down; e) all horizontal an vertical asymptotes. $$ f(x)=\frac{1}{2} x^{2 / 3}-x^{1 / 3} $$

Step-by-Step Solution

Answer

Function decreases on (0, 1), increases on (1, ∞); concave up (1, 8), concave down (8, ∞); local min at x=1, inflection at x=8; horizontal asymptote y=0.

Step 1: Identify critical points

To find critical points, we need to compute the derivative of the function and set it equal to zero. The function is \( f(x) = \frac{1}{2} x^{2/3} - x^{1/3} \). The derivative can be computed as follows: \[f'(x) = \frac{1}{2} \cdot \frac{2}{3} x^{-1/3} - \frac{1}{3} x^{-2/3} = \frac{1}{3}x^{-1/3} - \frac{1}{3}x^{-2/3}\]Setting \( f'(x) = 0 \) gives \( \frac{1}{3}x^{-1/3} - \frac{1}{3}x^{-2/3} = 0 \). Simplifying, we set \( x^{-1/3}(1 - x^{-1/3}) = 0 \) leading to \( x^{-1/3} = 0 \) which is undefined, or \( 1 - x^{-1/3} = 0 \), resulting in \( x = 1 \). Thus, the critical point is \( x = 1 \).

Step 2: Determine intervals of increase or decrease

Using \( x = 1 \) as a critical point, apply the first derivative test. Test intervals around \( x = 1 \) (for instance, \( x < 1 \) and \( x > 1 \)) to determine whether the function is increasing or decreasing. For \( x < 1 \), choose \( x = 0 \): \( f'(0) = \frac{1}{3}(0)^{-1/3} - \frac{1}{3}(0)^{-2/3} \) (consider a limit as \( x \to 0^+ \)) indicates decreasing behavior.For \( x > 1 \), choose \( x = 8 \): \( f'(8) = \frac{1}{3}(8)^{-1/3} - \frac{1}{3}(8)^{-2/3} = \frac{1}{6} - \frac{1}{12} = \frac{1}{12} \) indicates increasing behavior.Thus, the function is decreasing on \( (0, 1) \) and increasing on \( (1, \infty) \).

Step 3: Identify concavity intervals and inflection points

Compute the second derivative: \[f''(x) = -\frac{1}{9}x^{-4/3} + \frac{2}{9}x^{-5/3}\]Find where \( f''(x) = 0 \): \(-\frac{1}{9}x^{-4/3} + \frac{2}{9}x^{-5/3} = 0 \) leads to \( x^{-4/3}(2x^{-1/3} - 1) = 0 \). Thus, \( x = 8 \) (considering only positive roots).To evaluate concavity, test intervals around \( x = 8 \).For \( x < 8 \), \( f''(1) = -\frac{1}{9} + \frac{2}{9} = \frac{1}{9} > 0 \) shows concave up.For \( x > 8 \), \( f''(27) = -\frac{1}{9}\cdot 27^{-4/3} + \frac{2}{9}\cdot 27^{-5/3} \approx -0.015 \), shows concave down.Inflection point at \( x = 8 \).

Step 4: Check for asymptotes

To find asymptotes, consider limits. Horizontal asymptotes are found by evaluating \(\lim_{x \to \infty} \left(\frac{1}{2}x^{2/3} - x^{1/3}\right) = 0.\)Vertical asymptotes occur where the function is undefined, which isn’t the case here because both components are defined for all \( x > 0 \).

Step 5: Sketch the graph

Plot the function based on the information gathered:- Decreasing on \((0, 1)\), increasing on \((1, \infty)\)- Concave up on \((1, 8)\), concave down on \((8, \infty)\)- Critical points at \( x = 1 \) (local minimum), inflection point at \( x = 8 \)- Horizontal asymptote at \( y = 0 \).Combine the above insights to sketch a detailed graph.

Key Concepts

Critical points

Critical points are specific values of \( x \) where the derivative of the function \( f(x) \) is zero or undefined. These points are crucial as they can tell us a lot about the function's behavior, especially concerning local maximums and minimums. In the case of the function \( f(x) = \frac{1}{2} x^{2/3} - x^{1/3} \), the critical point arises when \( f'(x) = 0 \). Calculating the derivative gives:

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

By setting this to zero, we find it simplifies to:

• \( \frac{1}{3}x^{-1/3} - \frac{1}{3}x^{-2/3} = 0 \)

This leads us to the critical point at \( x = 1 \). This point is noteworthy because it's often where the graph changes direction from increasing to decreasing, or vice versa. Hence, critical points help identify potential peaks and troughs in the graph.

Concavity and inflection points

Concavity helps us understand the curvature of a function. It describes whether the graph bends upwards (concave up) or downwards (concave down). We check concavity by examining the second derivative \( f''(x) \). For our function, the second derivative is:

• \( f''(x) = -\frac{1}{9}x^{-4/3} + \frac{2}{9}x^{-5/3} \)

Inflection points occur where the concavity changes. Solving \( f''(x) = 0 \):

• This gives \( x = 8 \), indicating a switch in concavity at this point.

By evaluating \( f''(x) \) on intervals around \( x = 8 \), we see:

• For \( x < 8 \), \( f''(x) > 0 \), implying the function is concave up.
• For \( x > 8 \), \( f''(x) < 0 \), implying the function is concave down.

Thus, \( x = 8 \) is an inflection point, marking a transition in the graph's bending behavior. This understanding helps us sketch more accurate and visually representative graphs.

Asymptotes

Asymptotes are lines that the graph of a function approaches but never touches. They can be horizontal, vertical, or oblique. For the function \( f(x) = \frac{1}{2} x^{2/3} - x^{1/3} \), we particularly look for horizontal and vertical asymptotes. Horizontal asymptotes occur at limits as \( x \to \pm \infty \). Here:

• \( \lim_{x \to \infty} (\frac{1}{2}x^{2/3} - x^{1/3}) = 0 \)

This indicates a horizontal asymptote at \( y = 0 \). Vertical asymptotes occur where the function is undefined, often at points where the denominator is zero in a rational function. However, in this exercise, the function is defined for all \( x > 0 \) due to the real-valued powes, so there are no vertical asymptotes. Understanding these invisible boundaries helps predict the end behavior of a function's graph.

Increasing and decreasing intervals

The behavior of a graph over specific intervals, either increasing or decreasing, can be determined using the first derivative of the function. Recall that the derivative \( f'(x) \) informs us of these intervals:

• It is negative in regions where the function is decreasing.
• It is positive in regions where the function is increasing.

In our function, by examining \( f'(x) \), we determined:

• The function is decreasing on \((0, 1)\).
• The function is increasing on \((1, \infty)\).

Interpreting these intervals accurately allows us to detect the overall trend of the function over its domain. These insights reflect changes in direction, marked by critical points, that are valuable in sketching and understanding the graph's visual trends.