Problem 6 - College Algebra and Calculus: An Applied Approach

Question

In Exercises, find all relative extrema of the function. $$ g(x)=\frac{1}{5} x^{5}-x $$

Step-by-Step Solution

Answer

The function \(g(x)=\frac{1}{5} x^{5}-x\) has a local maximum at \(x = -1\) and a local minimum at \(x = 1\).

Step 1: Find the derivative

The first step is to compute the derivative of the function. The derivative of \(g(x) = \frac{1}{5} x^{5}-x\) is \(g'(x) = x^{4}-1\).

Step 2: Find critical points

The next step is to set the derivative equal to zero and solve for x, as the solutions will be the critical points. Hence, we set \(x^{4}-1 = 0\), which simplifies to \(x^{4} = 1\). The roots of this equation are \(x= 1, -1\), such that these are the critical points of \(g(x)\)

Step 3: Second Derivative Test

Next, we compute the second derivative to use the second derivative test. It is \(g''(x) = 4x^{3}\). We plug the critical points into the second derivative: \(g''(1) = 4 > 0\) and \(g''(-1) = -4 < 0\). This indicates that \(x = 1\) is a local minimum and \(x = -1\) is a local maximum.

Key Concepts

Critical Points

Critical points in a function are locations on the graph where the slope of the tangent is zero or undefined. Typically, these correspond to potential candidates for relative maxima, minima, or saddles. However, not all critical points guarantee an extrema. For the function given by:

• \(g(x) = \frac{1}{5} x^{5} - x\)

first, we located the critical points by finding where its derivative equals zero. The derivative,

• \(g'(x) = x^4 - 1\)

equals zero when

• \((x^4 - 1) = 0\).

Solving this gives us the critical points

• \(x = 1\) and \(x = -1\).

These points may potentially be locations where the function changes direction. But to make that determination, further investigation using derivative tests is needed.

First Derivative Test

The First Derivative Test helps determine whether a critical point is a local maximum, minimum, or neither by analyzing the sign changes of the first derivative before and after each critical point. In our function, we didn't utilize this test directly, but here's how it would work:

• Examine the sign of \(g'(x) = x^4 - 1\) around each critical point \(x = 1\) and \(x = -1\).
• For \(x = 1\):
• Just before \(x = 1\), at \(x = 0.9\), \(g'(x)\) is negative.
• Just after \(x = 1\), at \(x = 1.1\), \(g'(x)\) is positive.
• The derivative changes from negative to positive indicating a local minimum at \(x = 1\).
• For \(x = -1\):
• Just before \(x = -1\), at \(x = -1.1\), \(g'(x)\) is positive.
• Just after \(x = -1\), at \(x = -0.9\), \(g'(x)\) is negative.
• The derivative changes from positive to negative indicating a local maximum at \(x = -1\).

This method is insightful, especially when the Second Derivative Test might be inconclusive, providing clarity on changes in monotonicity.

Second Derivative Test

The Second Derivative Test offers a straightforward way to test critical points of a function for local maxima or minima by using the second derivative.

• The second derivative of our function:
  • \(g''(x) = 4x^3\)

is calculated by differentiating the first derivative. Once the second derivative has been found, it is evaluated at the critical points.

• At \(x = 1\), \(g''(1) = 4(1)^3 = 4\).
• Since \(g''(1) > 0\), it indicates a concave up curve at \(x = 1\), confirming a local minimum.
• At \(x = -1\), \(g''(-1) = 4(-1)^3 = -4\).
• Here, \(g''(-1) < 0\), indicating a concave down curve, therefore \(x = -1\) is a local maximum.

This test is often favored for its efficiency in assessing the nature of critical points when the derivative can be easily evaluated. The Second Derivative Test is usually reliable except in circumstances where the second derivative at the critical point is zero, thus requiring alternate methods like the First Derivative Test.