Problem 41 - Calculus Early Transcendentals

Question

Use any method to find the relative extrema of the function \(f .\) $$ f(x)=2 x+3 x^{2 / 3} $$

Step-by-Step Solution

Answer

Relative maximum at \(x = -1\) and relative minimum at \(x = 0\).

Step 1: Identify Critical Points

To find the critical points where the relative extrema could occur, we first need to find the derivative of the function. This is because critical points are where the derivative is zero or undefined. Given \( f(x) = 2x + 3x^{2/3} \), we differentiate it using the power rule. The derivative is:\[ f'(x) = 2 + 2x^{-1/3} \]Next, we set the derivative equal to zero to find the x-values of potential extrema:\[ 2 + 2x^{-1/3} = 0 \]Rearrange and solve for \( x \):\[ 2 = -2x^{-1/3} \]\[ 1 = -x^{-1/3} \]\[ x^{-1/3} = -1 \]Which simplifies to \( \ x^{1/3} = -1 \), hence \( x = -1 \). We also find where the derivative is undefined. Since the derivative contains \( x^{-1/3} \), it is undefined when \( x = 0 \). Thus, the critical points are \( x = -1 \) and \( x = 0 \).

Step 2: Test Critical Points

To determine if these critical points are relative maxima or minima, we use the second derivative test. First, find \( f''(x) \):\[ f''(x) = -\frac{2}{3}x^{-4/3} \]Now, apply the second derivative test:- At \( x = -1 \): \( f''(-1) = -\frac{2}{3}(-1)^{-4/3} = -\frac{2}{3} \), which is less than 0, indicating a relative maximum.- At \( x = 0 \): \( f''(0) \) is undefined, meaning we need a different approach here. For testing purposes, substitute values around \( x = 0 \) into the first derivative, showing a sign change from negative to positive, indicating a local minimum at \( x = 0 \).

Step 3: Conclusions

Summarize findings based on the tests. We have found that there is a relative maximum at \( x = -1 \) and a relative minimum at \( x = 0 \). Therefore, the function \( f(x) \) has a relative maximum at \( x = -1 \) and a relative minimum at \( x = 0 \).

Key Concepts

Critical Points

Critical points in a function are vital for finding where relative extrema, such as maxima or minima, might occur. These are the points on the graph of a function where the slope is zero or the derivative is undefined. Critically, it is where a function can change from increasing to decreasing, or vice versa.
To find the critical points of the function \( f(x) = 2x + 3x^{2/3} \), we begin by calculating its derivative. Applying the power rule gives us:

• \( f'(x) = 2 + 2x^{-1/3} \)

Here, a critical point exists where \( f'(x) = 0 \) or where the derivative does not exist.
Next, we solve \( 2 + 2x^{-1/3} = 0 \), rearranging to find when \( x^{1/3} = -1 \), thus \( x = -1 \). Another critical point is found where the function's derivative is undefined, which occurs at \( x = 0 \) due to the term \( x^{-1/3} \). Therefore, for this function, the critical points are \( x = -1 \) and \( x = 0 \).

Derivative Test

The derivative test helps to ascertain whether critical points are in fact relative maxima or minima. This involves looking at the behavior of the derivative before and after the critical points.
For \( f'(x) = 2 + 2x^{-1/3} \), we determine the sign changes around the critical points.

• At \( x = -1 \) and \( x = 0 \), check values just greater and less than these points.
• If \( f'(x) \) changes from positive to negative at a critical point, it indicates a relative maximum.
• If \( f'(x) \) changes from negative to positive, it shows a relative minimum.

Trying values around \( x = -1 \), \( f'(x) \) changes sign from positive to negative, confirming a relative maximum. Around \( x = 0 \), as the derivative changes from negative to positive, it indicates a local minimum.
Hence, the derivative test effectively helps to classify the nature of critical points.

Second Derivative Test

The second derivative test is a method used to classify critical points beyond the basic derivative test. Here, it determines the concavity of the function around the critical points.
Calculate the second derivative \( f''(x) = -\frac{2}{3}x^{-4/3} \). With it, we test each critical point:

• At \( x = -1 \), evaluate \( f''(-1) = -\frac{2}{3} \). The negative value confirms downward concavity, indicating a relative maximum at this point.
• At \( x = 0 \), \( f''(0) \) turns out to be undefined, suggesting a more nuanced approach is necessary. Here, assessing nearby points shows a change in \( f'(x) \) from negative to positive, revealing a local minimum.

The second derivative offers deeper insight into the curvature of the function at critical points, strengthening our understanding of their roles as relative extrema.