Problem 34 - Calculus Early Transcendentals

Question

Find the relative extrema using both first and second derivative tests. $$ f(x)=x^{4}-12 x^{3} $$

Step-by-Step Solution

Answer

The local maximum is at \( x = 0 \) and the local minimum is at \( x = 9 \).

Step 1: Find the First Derivative

To begin, calculate the first derivative of the function \( f(x) = x^4 - 12x^3 \). Using the power rule, the derivative is: \[ f'(x) = 4x^3 - 36x^2 \]

Step 2: Solve for Critical Points

Next, set the first derivative equal to zero to find critical points:\[ 4x^3 - 36x^2 = 0 \]Factor the equation:\[ 4x^2 (x - 9) = 0 \]Thus, the critical points are \( x = 0 \) and \( x = 9 \).

Step 3: First Derivative Test

To determine whether each critical point is a maximum or minimum, examine the sign of \( f'(x) \) around the critical points.1. For \( x = 0 \), test values: - If \( x = -1 \), \( f'(-1) > 0 \) (positive). - If \( x = 1 \), \( f'(1) < 0 \) (negative). - Thus, \( x = 0 \) is a local maximum.2. For \( x = 9 \), test values: - If \( x = 8 \), \( f'(8) < 0 \) (negative). - If \( x = 10 \), \( f'(10) > 0 \) (positive). - Thus, \( x = 9 \) is a local minimum.

Step 4: Find the Second Derivative

Find the second derivative of \( f(x) \).\[ f''(x) = 12x^2 - 72x \]

Step 5: Second Derivative Test

Evaluate the second derivative at the critical points to confirm extremum type.1. For \( x = 0 \):\[ f''(0) = 12(0)^2 - 72(0) = 0 \] The second derivative test is inconclusive. Use the first test result.2. For \( x = 9 \):\[ f''(9) = 12(9)^2 - 72(9) = 972 - 648 = 324 > 0 \] So, \( x = 9 \) is a local minimum.

Key Concepts

First Derivative Test

The First Derivative Test is an essential method used to find relative extrema, which are the local maxima or minima of a function. Here's a simple way to use this test:

• First, calculate the first derivative of the function. This gives the slope of the tangent line to the curve at any given point.
• Next, find the critical points by setting the first derivative equal to zero. These points are where the function might change direction, indicating potential maxima or minima.
• Finally, test intervals around each critical point. Check the sign of the first derivative just before and after each critical point to determine the nature of the extremum.
  • If the sign changes from positive to negative, you have a local maximum.
  • If the sign changes from negative to positive, you have a local minimum.

By using this step-by-step approach, you can determine whether each critical point is a peak or a dip in the function's graph.

Second Derivative Test

The Second Derivative Test offers another way to analyze and confirm the nature of critical points identified by the first derivative. This test involves evaluating the second derivative at critical points and is mainly used for:

• Validating whether a critical point is an extremum and determining its type.

Steps to apply the Second Derivative Test:

• First, find the second derivative of the function from which you already have the first derivative.
• Next, substitute the critical points into the second derivative.
• Analyze the result at each critical point:
  • If the second derivative is positive, the critical point is a local minimum (indicating the curve is concave up).
  • If the second derivative is negative, the critical point is a local maximum (indicating the curve is concave down).
  • If the second derivative is zero, the test is inconclusive, meaning you'll need the first derivative test for more information.

Overall, the second derivative provides insight into the curvature of the function, helping ascertain whether you have peaks or troughs.

Critical Points

Critical Points are pivotal in analyzing the behavior of functions, serving as candidates for relative extrema. They are places where the derivative is zero or undefined, indicating possible points of interest on the curve.

To identify critical points:

• Begin by finding the first derivative of your function. Critical points occur where this derivative equals zero or does not exist.
• Solve the derivative equation for these points to get the critical values of the function.

After you have these points, further testing is necessary to determine the nature of these critical points:

• Use the first derivative and second derivative tests to ascertain whether these points represent local maxima, minima, or if they are saddle points or points of inflection.

In our example, the critical points found were at the roots of the equation derived from the first derivative. These form the foundation for applying other tests to thoroughly explore the function's characteristics.