Problem 6 - Calculus Early Transcendentals

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 \(x=0\), but the nature of these points is inconclusive with the second derivative test; the first derivative test suggests \(x=0\) is an inflection point.

Step 1: Determine the first derivative of f(x)

To find the stationary points, we first need to determine the derivative of the function. For \( f(x) = 1 - x^5 \), the first derivative is \( f'(x) = -5x^4 \).

Step 2: Set the first derivative of f(x) to zero

Stationary points occur where the derivative is zero. Set \( f'(x) = -5x^4 = 0 \). Solving this equation gives \( x = 0 \).

Step 3: Determine the first derivative of g(x)

Next, find the derivative of \( g(x) = 3x^4 - 8x^3 \). The first derivative is \( g'(x) = 12x^3 - 24x^2 \).

Step 4: Set the first derivative of g(x) to zero

For stationary points, solve \( g'(x) = 12x^3 - 24x^2 = 0 \). Factor the equation as \( 12x^2(x - 2) = 0 \), giving solutions \( x = 0 \) and \( x = 2 \). Therefore, \( x = 0 \) is a stationary point.

Step 5: Find the second derivative of f(x)

The second derivative of \( f(x) = 1 - x^5 \) is \( f''(x) = -20x^3 \). Evaluate \( f''(x) \) at the stationary point: \( f''(0) = -20(0)^3 = 0 \), indicating the second derivative test is inconclusive at \( x = 0 \).

Step 6: Find the second derivative of g(x)

The second derivative of \( g(x) = 3x^4 - 8x^3 \) is \( g''(x) = 36x^2 - 48x \). Evaluate \( g''(x) \) at \( x = 0 \): \( g''(0) = 36(0)^2 - 48(0) = 0 \), indicating the second derivative test is inconclusive at \( x = 0 \).

Step 7: Discuss the first derivative test for f(x) and g(x)

Since the second derivative was inconclusive, use the first derivative test. For \( f(x) = 1 - x^5 \), analyze \( f'(x) = -5x^4 \). For \( g(x) = 3x^4 - 8x^3 \), analyze \( g'(x) = 12x^3 - 24x^2 \). Neither function's first derivative changes sign around \( x = 0 \), indicating neither a local maximum nor a local minimum, suggesting \( x = 0 \) is an inflection point for both functions.

Key Concepts

First Derivative

The first derivative of a function is important for finding its stationary points. These points are where the function's slope is zero, meaning it is neither increasing nor decreasing at that exact moment.
To find the first derivative, we differentiate the function.* For the function \( f(x) = 1 - x^5 \), the first derivative is \( f'(x) = -5x^4 \).
* For \( g(x) = 3x^4 - 8x^3 \), the first derivative is \( g'(x) = 12x^3 - 24x^2 \).Next, we determine the stationary points by setting the first derivatives to zero:

• \( f'(x) = 0 \) gives us \( x = 0 \), showing \( x = 0 \) is a stationary point for \( f(x) \).
• For \( g'(x) = 0 \), we factor the equation as \( 12x^2(x - 2) = 0 \), revealing solutions \( x = 0 \) and \( x = 2 \). Therefore, \( x = 0 \) is a stationary point for \( g(x) \).

Identifying stationary points through the first derivative helps assess where the function might change direction or rest.

Second Derivative

The second derivative provides further insights into the nature of stationary points. It helps us determine whether these points are potential maximums, minimums, or inflection points.
The second derivative is found by differentiating the first derivative once more.Here's how it works:

• The second derivative of \( f(x) = 1 - x^5 \) is \( f''(x) = -20x^3 \). Evaluating \( f''(0) \) yields \( 0 \), making the second derivative test inconclusive at \( x = 0 \) for \( f(x) \).
• Similarly, the second derivative for \( g(x) = 3x^4 - 8x^3 \) results in \( g''(x) = 36x^2 - 48x \). Evaluating at \( x = 0 \) gives \( 0 \), also rendering the second derivative test inconclusive for \( g(x) \) at \( x = 0 \).

When the second derivative at a stationary point equals zero, as seen here, it means we must use another method to determine the point's nature, such as the first derivative test.

Inflection Point

When stationary points have inconclusive second derivative tests, as with both \( f(x) \) and \( g(x) \) in this scenario, we often suspect an inflection point. An inflection point marks where the concavity of a function changes but does not necessarily indicate a local maximum or minimum.
To confirm inflection:

• For \( f(x) = 1 - x^5 \), the first derivative \( f'(x) = -5x^4 \) is always non-negative near \( x = 0 \). This lack of sign change indicates that \( x = 0 \) is an inflection point.
• Similarly, the first derivative \( g'(x) = 12x^3 - 24x^2 \) shows no sign change around \( x = 0 \). Thus, \( x = 0 \) is an inflection point in \( g(x) \) too.

Inflection points like these demonstrate subtle changes in the behavior of the curve, where neither extreme peaks nor troughs are present, yet some alteration in the rate of change occurs. Noticing this helps in understanding the shape and direction of a graph precisely.