Problem 5 - Calculus: Early Transcendentals

Question

Find all critical points and identify them as local maximum points, local minimum points, or neither. $$ y=3 x^{4}-4 x^{3} $$

Step-by-Step Solution

Answer

Critical points are \(x = 0\) (neither maximum nor minimum) and \(x = 1\) (local minimum).

Step 1: Find the First Derivative

To determine the critical points, start by finding the first derivative of the function. The given function is \[ y = 3x^4 - 4x^3. \] Compute the derivative using the power rule: \[ \frac{dy}{dx} = 12x^3 - 12x^2. \]

Step 2: Set the Derivative to Zero

Set the derivative equal to zero to find the critical points: \[ 12x^3 - 12x^2 = 0. \] Factor out the greatest common factor, which is \(12x^2\): \[ 12x^2(x - 1) = 0. \] This gives the critical points as \(x = 0\) and \(x = 1\).

Step 3: Find the Second Derivative

To classify the critical points, compute the second derivative of the original function. The first derivative is \[ 12x^3 - 12x^2. \] The second derivative is \[ \frac{d^2y}{dx^2} = 36x^2 - 24x. \]

Step 4: Evaluate the Second Derivative at Critical Points

Evaluate the second derivative at the critical points to determine their nature:For \(x = 0\):\[ \frac{d^2y}{dx^2} (0) = 36(0)^2 - 24(0) = 0. \]This test is inconclusive.For \(x = 1\):\[ \frac{d^2y}{dx^2} (1) = 36(1)^2 - 24(1) = 12. \]Since this is positive, \(x = 1\) is a local minimum.

Step 5: Use the First Derivative Test for Inconclusive Points

For the point \(x = 0\) where the second derivative test was inconclusive, use the first derivative test. The factorization of the first derivative was \[ 12x^2(x - 1). \]Choose test points around \(x = 0\), like \(x = -1\) and \(x = 0.5\):- For \(x = -1\), \(12(-1)^2(-1 - 1) = -24\) (negative).- For \(x = 0.5\), \(12(0.5)^2(0.5 - 1) = -1.5\) (negative).The sign does not change around zero, suggesting \(x = 0\) is not a local extremum.

Key Concepts

Local Maximum

A local maximum occurs at a critical point where the function value is higher than the values of the function at nearby points. With the first derivative at the critical point equaling zero, it suggests a potential peak or valley in the graph of the function.

• To identify if a critical point is a local maximum, the second derivative test can be used. This involves taking the second derivative of the function and evaluating it at the critical point.
• If the second derivative at this point is negative, it confirms that the critical point is a local maximum.

For example, in our given problem, we analyzed critical points obtained from the derivative but found that none corresponded to a local maximum. This is confirmed by applying the second derivative test, which showed no negative results for the critical points.

Local Minimum

The concept of a local minimum is crucial when analyzing a function to understand its behavior. A local minimum happens when a critical point has a function value lower than the immediate surrounding points.

• In practical terms, at a local minimum, the graph of the function experiences a "dip" or "valley".
• The second derivative test is helpful in confirming a local minimum. If the second derivative evaluated at a critical point is positive, the point is a local minimum.

In our exercise, we see this with the critical point at \(x = 1\). The second derivative here is positive at \(x = 1\), thus identifying it as a local minimum.

Second Derivative Test

The second derivative test is a powerful tool that helps determine the nature of critical points. Once the first derivative is set to zero to find critical points, the second derivative comes into play.

• The second derivative, denoted as \(\frac{d^2y}{dx^2}\), is simply the derivative of the first derivative.
• It provides information about the "concavity" of the function at the critical points.

If the second derivative at a critical point is positive, it indicates a local minimum. Conversely, if it's negative, it suggests a local maximum. A zero result means the test is inconclusive, while requiring further analysis, such as the first derivative test.

First Derivative

The first derivative represents the rate of change of the function. Calculating the first derivative is the initial step in finding critical points.

• By setting the first derivative equal to zero, you can solve for \(x\)-values that potentially represent the location of maxima, minima, or points of inflection.
• In our problem, the first derivative turned out to be \(12x^3 - 12x^2\), which we solved to find critical points \(x = 0\) and \(x = 1\).

Each of these critical points can then be analyzed with either the second derivative test or other means to determine their exact nature.

Power Rule

The power rule in calculus is a fundamental technique for finding derivatives. It simplifies computing derivatives of functions raised to a power.

• The power rule states that for a function \(f(x) = x^n\), its derivative \(f'(x) = nx^{n-1}\).
• This rule was applied to our problem to find the first derivative of \(y = 3x^4 - 4x^3\), resulting in \(12x^3 - 12x^2\).

By using this method, we can efficiently and accurately find the rate of change of polynomial functions, allowing us to explore their critical points and overall behavior.