Problem 6 - Calculus - AP Edition

Question

(a) Show that \(f(x)=1-x^{5}\) and \(g(x)=3 x^{4}-8 x^{3}\) both have stationary points at \(x=0\) (b) What does the second derivative test tell you about the nature of these stationary points? (c) What does the first derivative test tell you about the nature of these stationary points?

Step-by-Step Solution

Answer

Both functions have stationary points at \(x=0\); second derivative test is inconclusive, and the first derivative test shows \(f(x)\) is decreasing around \(x=0\), \(g(x)\) test is inconclusive.

Step 1: Find the First Derivative of f(x)

The function given is \( f(x) = 1 - x^5 \). To find the stationary points, we first need the derivative: \( f'(x) = \frac{d}{dx}(1 - x^5) = -5x^4 \).

Step 2: Solve f'(x) = 0 for Stationary Points

To find the stationary points of \( f(x) \), set \( f'(x) = -5x^4 = 0 \). Solving this gives \( x = 0 \). Thus, \( f(x) \) has a stationary point at \( x = 0 \).

Step 3: Find the First Derivative of g(x)

Now, consider the function \( g(x) = 3x^4 - 8x^3 \). The derivative is \( g'(x) = \frac{d}{dx}(3x^4 - 8x^3) = 12x^3 - 24x^2 \).

Step 4: Solve g'(x) = 0 for Stationary Points

To find the stationary points of \( g(x) \), set \( g'(x) = 12x^3 - 24x^2 = 0 \). Factoring gives \( 12x^2(x - 2) = 0 \), so \( x = 0 \) or \( x = 2 \). Thus, \( g(x) \) has a stationary point at \( x = 0 \).

Step 5: Find the Second Derivative of f(x)

To use the second derivative test, evaluate \( f''(x) = \frac{d}{dx}(-5x^4) = -20x^3 \). At \( x = 0 \), \( f''(0) = -20(0)^3 = 0 \). The second derivative test is inconclusive.

Step 6: Find the Second Derivative of g(x)

Evaluate \( g''(x) = \frac{d}{dx}(12x^3 - 24x^2) = 36x^2 - 48x \). At \( x = 0 \), \( g''(0) = 36(0)^2 - 48(0) = 0 \). The second derivative test is inconclusive.

Step 7: Apply the First Derivative Test to f(x) at x=0

To use the first derivative test, observe \( f'(x) = -5x^4 \) changes sign only at \( x = 0 \). Since \( -5x^4 \) is negative for all \( x eq 0 \), \( f(x) \) is decreasing around \( x=0 \). This suggests \( x=0 \) is neither a local minimum nor maximum.

Step 8: Apply the First Derivative Test to g(x) at x=0

For \( g(x) \), look at \( g'(x) = 12x^3 - 24x^2 \). Checking points around \( x=0 \), since \( g'(x) \) does not clearly change sign before crossing zero, the test is inconclusive at \( x=0 \).

Key Concepts

First Derivative

The first derivative of a function is a fundamental concept in calculus used to find stationary points, which are points where the slope of the tangent to the curve is zero.
The process involves differentiating the function with respect to its variable and setting the derivative equal to zero.
For the given functions, we begin by differentiating:

• For \( f(x) = 1 - x^5 \), the first derivative is \( f'(x) = -5x^4 \).
• For \( g(x) = 3x^4 - 8x^3 \), the derivative is \( g'(x) = 12x^3 - 24x^2 \).

At stationary points, the derivative is zero. Thus, solving for \( f'(x) = 0 \) and \( g'(x) = 0 \) tells us the possible locations of the stationary points.
Solving \( f'(x) = -5x^4 = 0 \) gives us \( x = 0 \), indicating a stationary point there. Similarly, solving \( g'(x) = 12x^3 - 24x^2 = 0 \) leads to factors \( x^2 (x - 2) = 0 \), providing possible stationary points at \( x = 0 \) and \( x = 2 \), but here \( x = 0 \) is considered for the context of the problem.

Second Derivative Test

The second derivative test is used to determine the nature of stationary points (whether they are local maxima, minima, or points of inflection).
Once you've found a stationary point using the first derivative, you take the second derivative of the function and evaluate it at the stationary point.

• If \( f''(x) > 0 \), the point is a local minimum.
• If \( f''(x) < 0 \), the point is a local maximum.
• If \( f''(x) = 0 \), the test is inconclusive, and further analysis is needed.

For the functions involved:

• The second derivative of \( f(x) = 1 - x^5 \) is \( f''(x) = -20x^3 \). Evaluating at \( x = 0 \), \( f''(0) = 0 \), indicating the test is inconclusive for \( f(x) \).
• For \( g(x) = 3x^4 - 8x^3 \), the second derivative is \( g''(x) = 36x^2 - 48x \). At \( x = 0 \), \( g''(0) = 0 \), so again, the second derivative test is inconclusive for \( g(x) \).

When the second derivative test is inconclusive, other methods, such as the first derivative test or higher order derivatives, might be necessary to analyze the point further.

First Derivative Test

The first derivative test is used as another way to understand the nature of stationary points when the second derivative test is inconclusive.
In this method, you analyze the sign of the first derivative, \( f'(x) \), around the stationary points to check whether the function changes from increasing to decreasing or vice versa.

• For \( f(x) = 1 - x^5 \), the derivative is \( f'(x) = -5x^4 \), which is non-positive for all x. At \( x = 0 \), there is no change in the sign around it, meaning the point is neither a maximum nor a minimum.
• For \( g(x) = 3x^4 - 8x^3 \), the first derivative is \( g'(x) = 12x^3 - 24x^2 \). Around \( x = 0 \), we need to check the sign on small intervals around zero. Since the derivative doesn't clearly switch signs as it approaches and departs zero, the first derivative test at this point is also inconclusive.

The first derivative test can be particularly helpful in cases where the second derivative provides no clear answer on its own, offering another angle to approach the analysis by examining how the curve behaves.