Problem 25 - Calculus Early Transcendentals

Question

Use the given derivative to find all critical points of \(f\) and at each critical point determine whether a relative maximum, relative minimum, or neither occurs. Assume in each case that \(f\) is continuous everywhere. $$ f^{\prime}(x)=x^{2}\left(x^{3}-5\right) $$

Step-by-Step Solution

Answer

Critical points are at 0 (max) and \(\sqrt[3]{5}\) (min).

Step 1: Find Critical Points

Critical points occur when the derivative is zero or undefined. Our given derivative is \( f'(x) = x^2(x^3-5) \). Set the derivative equal to zero to find the critical points:\[f'(x) = x^2(x^3-5) = 0\]To solve this, we break it into two parts:- \( x^2 = 0 \), giving \( x = 0 \).- \( x^3 - 5 = 0 \), which gives \( x = \sqrt[3]{5} \).Thus, the critical points are \( x = 0 \) and \( x = \sqrt[3]{5} \).

Step 2: Determine Nature of Critical Points with Second Derivative Test

To determine whether each critical point is a relative maximum, relative minimum, or neither, we find the second derivative \( f''(x) \).Calculate \( f''(x) \):Start by expanding \( f'(x) = x^2(x^3-5) \):- Use the product rule: \( u = x^2 \), \( v = x^3 - 5 \).- \( u' = 2x \)- \( v' = 3x^2 \)Then,\[(u'v + uv') = 2x(x^3-5) + x^2(3x^2) = 2x^4 - 10x + 3x^4 = 5x^4 - 10x\]Thus, the second derivative is:\[f''(x) = 20x^3 - 10\]Evaluate \( f''(x) \) at each critical point:- At \( x = 0 \): \[f''(0) = 20(0)^3 - 10 = -10 < 0\] Therefore, \( x = 0 \) is a relative maximum.- At \( x = \sqrt[3]{5} \): \[f''(\sqrt[3]{5}) = 20(\sqrt[3]{5})^3 - 10 = 100 - 10 = 90 > 0\] Therefore, \( x = \sqrt[3]{5} \) is a relative minimum.

Key Concepts

First Derivative Test

The first derivative test is an essential tool in calculus used to analyze the behavior of functions. It helps determine where functions increase or decrease and identify relative extrema. First, it's necessary to find the critical points by setting the derivative of the function to zero and solving for the variable. In our example, the derivative given is \( f'(x) = x^2(x^3 - 5) \).
To find the critical points:

• Set the derivative equal to zero: \( f'(x) = 0 \).
• Solve \( x^2 = 0 \) which yields \( x = 0 \).
• Solve \( x^3 - 5 = 0 \) which gives \( x = \sqrt[3]{5} \).

These are the locations where the slope could change, and thus require further analysis to check if they indicate relative maxima or minima.

Second Derivative Test

The second derivative test is applied to determine the nature of critical points found using the first derivative. It can help establish if these points are relative maxima, minima, or neither.
Here’s how it works:

• Find the second derivative of the function, \( f''(x) \).
• In our case, the second derivative is \( f''(x) = 20x^3 - 10 \).
• Evaluate \( f''(x) \) at each critical point.
• If \( f''(x) > 0 \), the point is a relative minimum; if \( f''(x) < 0 \), it's a relative maximum; if \( f''(x) = 0 \), the test is inconclusive.

For \( x = 0 \):- \( f''(0) = -10 \); Since it’s less than zero, \( x = 0 \) is a relative maximum.
For \( x = \sqrt[3]{5} \):- \( f''(\sqrt[3]{5}) = 90 \); Since it’s greater than zero, \( x = \sqrt[3]{5} \) is a relative minimum.

Relative Extrema

In calculus, relative extrema refer to the points on a graph where the function value is a peak or valley compared to points close to it. These are known as relative maximum and minimum, occurring at critical points identified through tests like the first and second derivative tests.
- **Relative Maximum**: A point where the function value is higher than those nearby. For instance, at \( x = 0 \) in our exercise, the function has a local high.- **Relative Minimum**: A point where the function value is lower compared to its surroundings, like \( x = \sqrt[3]{5} \) in the example, indicating a local low.
Understanding relative extrema is crucial as they provide insights into the behavior of the function—where it rises or falls, assisting in drawing accurate graphs and interpreting real-world phenomena modeled by the function.