Problem 35 - Calculus Early Transcendentals

Question

Give a graph of the function and identify the locations of all critical points and inflection points. Check your work with a graphing utility. $$ 4 x^{1 / 3}-x^{4 / 3} $$

Step-by-Step Solution

Answer

Critical points: \(x=0\), \(x=1\). Inflection point: \(x=-2\).

Step 1: Find the First Derivative

First, find the derivative of the given function with respect to \( x \). The function is \( f(x) = 4x^{1/3} - x^{4/3} \).Using the power rule for derivatives, the derivative \( f'(x) \) is calculated as follows:\[ f'(x) = \frac{d}{dx}(4x^{1/3}) - \frac{d}{dx}(x^{4/3}) \]\[ f'(x) = 4 \cdot \frac{1}{3} x^{1/3 - 1} - \frac{4}{3} x^{4/3 - 1} \]\[ f'(x) = \frac{4}{3} x^{-2/3} - \frac{4}{3} x^{1/3} \]

Step 2: Find Critical Points

Critical points occur where the derivative is zero or undefined. Set \( f'(x) = 0 \) and solve for \( x \).\[ \frac{4}{3} x^{-2/3} - \frac{4}{3} x^{1/3} = 0 \]This simplifies to:\[ x^{-2/3} = x^{1/3} \]Multiply through by \( x^{2/3} \):\[ 1 = x \]So, \( x = 1 \) is a critical point.Also, consider where \( f'(x) \) is undefined: \( x^{-2/3} \) is undefined only when \( x = 0 \). So, \( x = 0 \) is a critical point as the derivative is undefined at this point.

Step 3: Find the Second Derivative

To find possible inflection points, compute the second derivative \( f''(x) \).\[ f''(x) = \frac{d}{dx} \left( \frac{4}{3} x^{-2/3} - \frac{4}{3} x^{1/3} \right) \]Using the power rule:\[ f''(x) = \frac{4}{3} \cdot (-\frac{2}{3}) x^{-2/3 - 1} - \frac{4}{3} \cdot \frac{1}{3} x^{1/3 - 1} \]\[ f''(x) = -\frac{8}{9} x^{-5/3} - \frac{4}{9} x^{-2/3} \]

Step 4: Find Inflection Points

Set \( f''(x) = 0 \) and solve for \( x \):\[ -\frac{8}{9} x^{-5/3} - \frac{4}{9} x^{-2/3} = 0 \]This simplifies to:\[ -8x^{-5/3} = 4x^{-2/3} \]Divide both sides by \( x^{-5/3} \):\[ -8 = 4x \]\[ x = -2 \]Thus, the inflection point is at \( x = -2 \). The inflection also occurs at any point where \( f''(x) \) is undefined (i.e., \( x = 0 \)), however, it must typically show a change in concavity to be an inflection point.

Step 5: Sketch the Graph

To confirm, plot the function \( f(x) = 4x^{1/3} - x^{4/3} \) and check identified critical points \( x = 1 \), \( x = 0 \), and inflection point \( x = -2 \).- At \( x = 1 \), observe a local extremum.- At \( x = 0 \), there is a vertical tangent.- At \( x = -2 \), verify a change in concavity visually.

Step 6: Verify with Graphing Utility

Use a graphing calculator or software (like Desmos or GeoGebra) to plot the function and confirm the locations of critical points at \( x = 0 \) and \( x = 1 \), and inflection at \( x = -2 \). Ensure the graph aligns with the mathematical calculations.

Key Concepts

Critical points

Critical points are very significant in the world of calculus when you're analyzing the graph of a function. These points occur where the function's derivative (slope) is either zero or fails to exist. At these locations, the graph typically experiences interesting behavior, such as turning points or peaks and valleys. Let's look at how to find critical points:

• First, you need to compute the first derivative of the function, denoted as \( f'(x) \).
• By setting \( f'(x) = 0 \), you determine where the slope of the tangent to the graph becomes horizontal.
• Additionally, points where the derivative is undefined are also candidates for critical points.

In the example function \( f(x) = 4x^{1/3} - x^{4/3} \), we derived the first derivative \( f'(x) = \frac{4}{3} x^{-2/3} - \frac{4}{3} x^{1/3} \). By setting it equal to zero, we found that \( x = 1 \) is a critical point. Likewise, since the derivative is undefined at \( x = 0 \), it's also considered a critical point.

Identifying these points can help predict where the graph could change direction, or where sharp turns exist.

Inflection points

Inflection points are intriguing features found on the graph of a function where the curvature changes direction. Unlike critical points where the slope is zero, inflection points involve the second derivative \( f''(x) \):

• The second derivative tells us about the rate of change of the slope – essentially, the concavity of the function.
• Inflection points occur where \( f''(x) = 0 \) or where the second derivative is undefined, but only if there is a visible change in concavity.

In our example with the function \( f(x) = 4x^{1/3} - x^{4/3} \), we calculated the second derivative as \( f''(x) = -\frac{8}{9} x^{-5/3} - \frac{4}{9} x^{-2/3} \). Solving for where the second derivative equals zero, we find an inflection point at \( x = -2 \). Additionally, \( f''(x) \) is undefined at \( x = 0 \); however, this point must display a change in concavity to classify as an inflection point.

Detecting these areas helps understand more nuanced transitions in a graph, revealing shifts in the function's growth pattern.

Derivatives

Derivatives are fundamental tools in calculus that help us understand the behavior of functions. They represent the rate at which a function is changing at any given point. Here's what you need to know about derivatives and how they relate to examining graphs:

• The first derivative \( f'(x) \) provides insights into the slope of the function's graph. It tells us whether the function is increasing or decreasing.
• Setting \( f'(x) = 0 \) helps identify critical points, which are potential extremums (maximum or minimum points) on the graph.
• The second derivative \( f''(x) \) gives information about the concavity of the function. It helps us detect inflection points.

For instance, with the function \( f(x) = 4x^{1/3} - x^{4/3} \), knowing \( f'(x) \) as \( \frac{4}{3} x^{-2/3} - \frac{4}{3} x^{1/3} \) allows us to find where the graph is stationary or changing quickly. Using \( f''(x) \), we observe concave upward or downward trends, thereby identifying points where these trends switch.

Understanding derivatives allows us to graph complex functions by breaking them down into useful pieces of information about their slope and curvature.