Problem 21 - Calculus: Early Transcendentals

Question

Find all local maximum and minimum points by the second derivative test. $$ y=3 x^{4}-4 x^{3} $$

Step-by-Step Solution

Answer

The function has a local minimum at \( x = 1 \), and the test is inconclusive at \( x = 0 \).

Step 1: Find the First Derivative

To find the critical points, we first need to find the first derivative of the function. Given the function \[ y = 3x^4 - 4x^3 \]we differentiate it with respect to \( x \):\[ y' = \frac{d}{dx}(3x^4 - 4x^3) = 12x^3 - 12x^2 \].

Step 2: Find the Critical Points

After obtaining the first derivative, we set it to zero to find the critical points:\[ 12x^3 - 12x^2 = 0 \]Factor out the greatest common factor:\[ 12x^2(x - 1) = 0 \]Setting each factor to zero, we have:1. \( 12x^2 = 0 \) leading to \( x = 0 \) 2. \( x - 1 = 0 \) leading to \( x = 1 \)Thus, the critical points are \( x = 0 \) and \( x = 1 \).

Step 3: Find the Second Derivative

We now find the second derivative to use it in the second derivative test:Given the first derivative \( y' = 12x^3 - 12x^2 \), we differentiate again:\[ y'' = \frac{d}{dx}(12x^3 - 12x^2) = 36x^2 - 24x \].

Step 4: Second Derivative Test Evaluation

Apply the second derivative test to each critical point.- For \( x = 0 \): Evaluate \( y''(0) \): \[ y''(0) = 36(0)^2 - 24(0) = 0 \] The test is inconclusive because the second derivative is zero.- For \( x = 1 \): Evaluate \( y''(1) \): \[ y''(1) = 36(1)^2 - 24(1) = 36 - 24 = 12 \] Since \( y''(1) > 0 \), the point \( x = 1 \) is a local minimum.

Step 5: Conclusion of Local Extremes

From the second derivative test, we conclude:1. The test is inconclusive for \( x = 0 \), implying we need further tests or visual inspection to confirm its nature.2. \( x = 1 \) is a local minimum since \( y''(1) > 0 \).

Key Concepts

Local Maximum

In calculus, identifying a local maximum helps understand where a function's graph peaks in a small surrounding area. A local maximum occurs at a point if the function value is higher than at nearby points. To determine these points, we use derivatives.
First, we find the function's first derivative (\( y' \)). Critical points where potential maxima occur are found when this derivative equals zero or is undefined.
After finding critical points, the second derivative (\( y'' \)) helps check these points for local extrema. Specifically, a point is a local maximum if:

• The second derivative at that point is negative (\( y''(x) < 0 \)).

It's essential to note that if the critical point has a zero second derivative, it may not be conclusive, leading to the need for other methods like plotting or higher-order tests to confirm the maximum.

Local Minimum

Understanding local minima in calculus helps identify where a function's graph dips, indicating a valley. A local minimum occurs when the function value is smaller than that at nearby points.
To ascertain these, first find the first derivative and solve to determine critical points. A function's local minimum at a critical point occurs if the second derivative is positive (\( y''(x) > 0 \)).
Thus:

• Determine critical points where \( y' = 0 \) or is undefined.
• Use the second derivative (\( y'' \)) to evaluate at each critical point.
• A positive second derivative indicates a local minimum.

In the given exercise, \( x = 1 \) with \( y''(1) = 12 \) indicates a local minimum, confirming the function is curving upwards.

Critical Points

Critical points play a vital role in finding where a function might have a local maximum, minimum, or neither. These occur where the function's derivative is zero or doesn't exist.
Determining these points involves:

• Calculating the first derivative, as seen in our example (\( y' = 12x^3 - 12x^2 \)).
• Solving \( y' = 0 \) to find the values of \( x \) where potential extrema might occur.

In our exercise, solving \( 12x^3 - 12x^2 = 0 \) resulted in critical points \( x = 0 \) and \( x = 1 \). These points are the foundation for further testing to determine their nature, using the second derivative test or alternative methods.

Second Derivative

The second derivative provides insights into the curvature or concavity of a function's graph. It plays a crucial role in the Second Derivative Test, which is used to classify the nature of critical points.
After finding a function's first derivative (\( y' \)), differentiating it gives the second derivative (\( y'' \)). The outcome of evaluating \( y'' \) at critical points determines whether there’s a local extremum:

• Positive \( y'' \) indicates a local minimum, meaning the graph is concave up.
• Negative \( y'' \) suggests a local maximum, with the graph being concave down.
• If \( y'' = 0 \), the test is inconclusive, and further analysis is required.

In our problem, the second derivative was found to be \( y'' = 36x^2 - 24x \), guiding the test: \( y''(0) = 0 \) (inconclusive) and \( y''(1) = 12 \) (local minimum). It tells us how sharply the graph increases or decreases at those points.