Problem 36 - 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. $$ f(x)=\left|x^{1 / 3}-4\right| $$

Step-by-Step Solution

Answer

Global minimum at \(x=0\); increasing for \(x\geq 64\); decreasing for \(x<64\); concave up for \(x<64\), concave down for \(x\geq 64\); no asymptotes.

Step 1: Identify Critical Points

To find local and global extrema, we need to identify the critical points of the function. Begin by writing the piecewise components for \( f(x) = |x^{1/3} - 4| \):\[ f(x) = \begin{cases} x^{1/3} - 4, & x \geq 64 \ -(x^{1/3} - 4), & x < 64 \end{cases} \] Determine where the derivative is zero or undefined.

Step 2: Find the Derivative

Differentiate the function piecewise: For \(x \geq 64\), \( f'(x) = \frac{1}{3}x^{-2/3} \); for \(x < 64\), \( f'(x) = -\frac{1}{3}x^{-2/3} \). Check where each derivative equals zero or is undefined. Here, the derivative does not equal zero, and it is undefined at \(x = 0\).

Step 3: Determine Extrema

Evaluate \( f(x) \) at critical points and endpoints. \( f(0) = |0^{1/3} - 4| = 4 \) is a critical point and is also the global minimum. There are no local extrema since the derivative does not change sign.

Step 4: Identify Inflection Points

Find the second derivative for concavity analysis. For \(x \geq 64\), \( f''(x) = -\frac{2}{9} x^{-5/3} \); for \(x < 64\), \( f''(x) = \frac{2}{9} x^{-5/3} \). Check for sign changes to find inflection points. No points result in a change in concavity.

Step 5: Determine Intervals of Increase/Decrease

Check the sign of \( f'(x) \) in the derived intervals. For \(x \geq 64\), \( f'(x) > 0 \) indicating increasing; for \(x < 64\), \( f'(x) < 0 \) indicating decreasing.

Step 6: Determine Concavity Intervals

Check the sign of \( f''(x) \). For \(x \geq 64\), \( f''(x) < 0 \), so the interval is concave down; for \(x < 64\), \( f''(x) > 0 \) indicating concave up.

Step 7: Analyze Asymptotes

The function has no vertical or horizontal asymptotes since it is defined for all \(x\) and approaches infinity as \(x\) approaches infinity.

Step 8: Plot the Graph

Using the intervals for increase, decrease, concavity and extremum points, sketch the graph of \( f(x) \). Label critical points, the minimum at \(x=0\), and the intervals of concavity and monotonicity.

Key Concepts

Local and Global Extrema

When sketching the graph of a function, understanding the extrema is crucial. Extremum points are where the function reaches local or global highs and lows. For the function\(f(x) = |x^{1/3} - 4|\), we are particularly interested in these points because they define the graph's peaks and valleys.

Local Extrema:

• These occur where the function changes direction locally. However, for our function, there are no local minima or maxima because the derivative does not change sign.

Global Extrema:

• A global minimum or maximum is the absolute smallest or largest value in the function's entire domain. In this case, the global minimum is at \(x=0\) with a value of 4.

Understanding extrema helps in plotting how the function behaves over its entire range, showing where it reaches its lowest and highest values.

Critical Points

Critical points are vital in analyzing a function's behavior. These points occur where the derivative is zero or undefined. For the function\(f(x) = |x^{1/3} - 4|\), identifying critical points involves calculating where its derivative equals zero or is not defined.

Finding Critical Points:

• For \(x \geq 64\), the derivative is \(\frac{1}{3}x^{-2/3}\), and for \(x < 64\), it is \(-\frac{1}{3}x^{-2/3}\).
• The derivative is never zero in the defined ranges but is undefined at \(x = 0\).
• This point, \(x = 0\), is significant as it corresponds to the global minimum of the function.

Recognizing critical points aids in determining intervals of increase and decrease in a graph, providing insight into the function's behavior.

Concavity

Concavity reveals how the slope of a function changes. It tells us whether a graph curves upwards or downwards. For our function, examining its concavity helps in understanding the graph's shape. Consider the second derivative for this analysis.

Concavity Analysis:

• For \(x \geq 64\), the second derivative is \(-\frac{2}{9} x^{-5/3}\), which is always negative, indicating concave down.
• For \(x < 64\), the second derivative is \(\frac{2}{9} x^{-5/3}\), always positive, indicating concave up.

Understanding concavity allows you to predict where the curve of the graph arches inwards or outwards, offering a clear picture of the function's overall shape over different intervals.

Inflection Points

Inflection points are where a graph changes its concavity. These points indicate a transition from concave up to concave down or vice versa. Checking the second derivative's sign changes helps in finding these points.

Checking for Inflection Points:

• For the function \(f(x) = |x^{1/3} - 4|\), the second derivative does not experience sign changes, indicating no inflection points.

Without inflection points, the graph maintains its concavity direction consistently over its intervals. While inflection points are essential for many functions to foresee changes in curvature, this particular function keeps a steady curve direction defined by its intervals.