Problem 11 - Essential Calculus Early Transcendentals

Question

Find the local maximum and minimum values of \(f\) using both the First and Second Derivative Tests. Which method do you prefer? \(f(x)=1+3 x^{2}-2 x^{3}\)

Step-by-Step Solution

Answer

Local max at \(x=1\), local min at \(x=0\). Both tests are useful, but preference depends on context; often the First Derivative Test is more intuitive.

Step 1: Find the First Derivative

To find the local maximum and minimum, we first calculate the first derivative of the function to determine the critical points. The given function is \( f(x) = 1 + 3x^2 - 2x^3 \). Let's differentiate it: \[ f'(x) = \frac{d}{dx}(1 + 3x^2 - 2x^3) = 6x - 6x^2 \] The first derivative is \( f'(x) = 6x - 6x^2 \).

Step 2: Find Critical Points

Critical points occur where the first derivative is zero or undefined. Set the first derivative to zero and solve for \( x \): \[ 6x - 6x^2 = 0 \] Factorizing gives: \[ 6x(1 - x) = 0 \] Thus, the critical points are \( x = 0 \) and \( x = 1 \).

Step 3: Apply the First Derivative Test

The First Derivative Test involves analyzing the sign changes of the first derivative around the critical points to determine if they are local maxima or minima:1. **Interval \((-\infty, 0)\):** Pick \(x = -1\). Substitute into \(f'(x) = 6x - 6x^2\) to get \(6(-1) - 6(-1)^2 = -6 - 6 = -12\) (Negative).2. **Interval \((0, 1)\):** Pick \(x = 0.5\). Substitute into \(f'(x)\) to get \(6(0.5) - 6(0.5)^2 = 3 - 1.5 = 1.5\) (Positive).3. **Interval \((1, \infty)\):** Pick \(x = 2\). Substitute into \(f'(x)\) to get \(6(2) - 6(2)^2 = 12 - 24 = -12\) (Negative).Based on sign changes:- At \(x=0\), the derivative changes from negative to positive, indicating a local minimum.- At \(x=1\), the derivative changes from positive to negative, indicating a local maximum.

Step 4: Find the Second Derivative

To confirm using the Second Derivative Test, we calculate the second derivative of \( f(x) \):\[ f''(x) = \frac{d}{dx}(6x - 6x^2) = 6 - 12x \]

Step 5: Apply the Second Derivative Test

The Second Derivative Test helps identify the nature of the critical points:1. **At \(x = 0\):** Substitute \(x=0\) into \(f''(x) = 6 - 12x\), yielding \(f''(0) = 6\) (Positive). Thus, \(x=0\) is a local minimum.2. **At \(x = 1\):** Substitute \(x=1\) into \(f''(x)\), yielding \(f''(1) = 6 - 12 = -6\) (Negative). Thus, \(x=1\) is a local maximum.

Key Concepts

First Derivative Test

The First Derivative Test is a method used in calculus to determine the local maxima and minima of a function using its first derivative. The test involves the following steps:

• First, find the critical points by setting the first derivative of the function, denoted as \( f'(x) \), equal to zero, and solving for \( x \). This step identifies where the function has potential peaks (maxima) or troughs (minima).
• Next, examine the intervals around these critical points. This is done by selecting test points in each interval created by the critical points.
• Evaluate the sign of \( f'(x) \) at these test points. By observing the sign change, you can infer the behavior of the function:
  • If \( f'(x) \) changes from negative to positive, the function transitions from decreasing to increasing, indicating a local minimum.
  • If \( f'(x) \) changes from positive to negative, the function transitions from increasing to decreasing, indicating a local maximum.

This test provides a clear visual of how the function behaves around its critical points, making it a simple and effective method.

Second Derivative Test

The Second Derivative Test is another useful tool in calculus for determining the nature of the critical points of a function. This method depends on the function's second derivative, \( f''(x) \). Here’s how it works:

• First, identify the critical points by setting \( f'(x) = 0 \) and solve for \( x \). These critical points are potential locations for local maxima or minima.
• Next, find the second derivative of the function, \( f''(x) \).
• Evaluate \( f''(x) \) at each critical point:
  • If \( f''(x) > 0 \), the function is concave up at that point, indicating a local minimum.
  • If \( f''(x) < 0 \), the function is concave down, indicating a local maximum.
  • If \( f''(x) = 0 \), the test is inconclusive and you might need to use other methods to determine the nature of that point.

While the Second Derivative Test can be easier and faster, it relies on the assumption of concavity via the second derivative, which may not always provide conclusive results.

Critical Points

In calculus, critical points are crucial in analyzing the behavior of functions. Critical points occur where the derivative of the function is either zero or undefined. These points are often where the function's behavior changes, such as peaks, troughs, or points of inflection.Here is a simplified guide to finding and analyzing critical points:

• Calculate the first derivative \( f'(x) \).
• Solve \( f'(x) = 0 \) for \( x \) to find potential critical points. If \( f'(x) \) does not exist at certain points, these are also critical points.
• Once identified, use tests like the First or Second Derivative Test to determine if these critical points are local maxima, minima, or saddle points.

Understanding critical points is essential as they provide insight into where and how the function changes direction, impacting overall function analysis.