Problem 5 - Calculus Early Transcendentals

Question

(a) Show that both of the functions \(f(x)=(x-1)^{4}\) and \(g(x)=x^{3}-3 x^{2}+3 x-2\) have stationary points at \(x=1 .\) (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

(a) Both have stationary points at \(x=1\). (b) Second derivative test is inconclusive. (c) \(f(x)\) has a local minimum, \(g(x)\) is a point of inflection.

Step 1: Find the first derivatives

Start by differentiating each function with respect to \(x\). For \(f(x) = (x-1)^4\), use the chain rule: \[ f'(x) = 4(x-1)^3. \] For \(g(x) = x^3 - 3x^2 + 3x - 2\), differentiate term by term: \[ g'(x) = 3x^2 - 6x + 3. \]

Step 2: Solve for stationary points

Set each first derivative to zero to find the stationary points. For \(f'(x) = 4(x-1)^3 = 0\), solve \((x-1)^3 = 0\), giving \(x=1\). For \(g'(x) = 3x^2 - 6x + 3 = 0\), factor as \(3(x^2 - 2x + 1) = 0\), which simplifies to \((x-1)^2 = 0\), giving again \(x=1\).

Step 3: Apply the second derivative test to f(x)

Find the second derivative of \(f(x)\), which is \(f''(x) = 12(x-1)^2\). Evaluate \(f''(1)\), which is \(12(1-1)^2 = 0\). The second derivative test is inconclusive because it equals zero.

Step 4: Apply the second derivative test to g(x)

Find the second derivative of \(g(x)\), which is \(g''(x) = 6x - 6\). Evaluate \(g''(1)\), which is \(6(1) - 6 = 0\). The second derivative test is also inconclusive here because it equals zero.

Step 5: Apply the first derivative test to f(x)

Consider intervals around \(x=1\), such as \((0.9, 1.1)\), and observe the sign of \(f'(x) = 4(x-1)^3\). For \(x<1\), \(f'(x)<0\), and for \(x>1\), \(f'(x)>0\). Therefore, \(x=1\) is a local minimum by the first derivative test.

Step 6: Apply the first derivative test to g(x)

Again, consider intervals such as \((0.9, 1.1)\), and observe the sign of \(g'(x) = 3(x-1)^2\). Since \((x-1)^2\) is always non-negative and \(g'(x)=0\) on the interval, \(x=1\) is a point of inflection, not a minimum or maximum.

Key Concepts

First Derivative Test

The First Derivative Test is a powerful tool in calculus for determining the nature of stationary points of a function. Stationary points occur where the first derivative of a function equals zero, meaning the function's slope is flat at that point.
To apply the First Derivative Test, we look at the sign of the first derivative before and after the point in question.

• If the derivative changes from positive to negative, the point is a local maximum.
• If it changes from negative to positive, the point is a local minimum.
• If there is no sign change, the point might be an inflection point.

In our original exercise, we found that for the function \(f(x) = (x-1)^4\), the derivative changes from negative to positive around \(x=1\). Thus, \(x=1\) is a local minimum.
For \(g(x) = x^3 - 3x^2 + 3x - 2\), the derivative does not change sign around \(x=1\). Instead, it remains zero, indicating that \(x=1\) is a point of inflection, not a maximum or minimum. This shows how the test helps categorize stationary points.

Second Derivative Test

The Second Derivative Test is a complementary method used in calculus to confirm the concavity of the function at stationary points and determine whether these points are local minima, maxima, or neither. It involves taking the second derivative of a function.
The test follows simple rules:

• If the second derivative is positive at the stationary point, the function is concave up, suggesting it's a local minimum.
• If the second derivative is negative, the function is concave down, indicating a local maximum.
• If the second derivative is zero, the test is inconclusive, and you may need other methods like the First Derivative Test.

In the exercise with \(f(x) = (x-1)^4\), we calculated the second derivative to be \(f''(1) = 0\), which is inconclusive. This means we cannot determine the type of stationary point using just the second derivative. Similarly, for \(g(x) = x^3 - 3x^2 + 3x - 2\), the second derivative at \(x=1\) was also zero, indicating that further investigation is required to understand the point's nature.

Calculus

Calculus is a branch of mathematics that deals with continuous change. It is divided into two main parts: differential calculus and integral calculus. Differential calculus focuses on the concept of a derivative, which represents change and how quantities differ. This is pivotal in understanding rates of change and slopes of curves.
In problems involving stationary points, as seen in the exercise, calculus allows us to:

• Determine where a function's derivative is zero (finding stationary points).
• Analyse the nature of these points using methods like the First and Second Derivative Tests.
• Explore function behavior near these points to correctly identify whether they are points of inflection, minima, or maxima.

By understanding these principles of calculus, students can identify the implications of stationary points in various mathematical contexts and how these points reflect the underlying behavior of the function in question. This solidifies why calculus is an essential tool in mathematics and science.