Question

Use the Taylor series about \(x=a\) to verify the familiar "second derivative test" for a maximum or minimum point. That is, show that if \(f^{\prime}(a)=0,\) then \(f^{\prime \prime}(a)>0\) implies a minimum point at \(x=a\) and \(f^{\prime \prime}(a)<0\) implies a maximum point at \(x=a\). Hint: For a minimum point, say, you must show that \(f(x)>f(a)\) for all \(x\) near enough to \(a\).

Short answer

If \(f'(a) = 0\) and \(f''(a) > 0\), then local minimum; if \(f''(a) < 0\), then local maximum.

Step-by-step solution

Express the Taylor Series

Start by expressing the function using the Taylor series expansion about the point \(x = a\). The general form of the Taylor series is given by: \[ f(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f'''(a)}{3!}(x - a)^3 + \cdots \]

Apply the Condition

Given that \(f'(a) = 0\), substitute it into the Taylor series expansion: \[ f(x) = f(a) + 0 \cdot (x - a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f'''(a)}{3!}(x - a)^3 + \cdots \] This simplifies to: \[ f(x) = f(a) + \frac{f''(a)}{2}(x - a)^2 + \frac{f'''(a)}{3!}(x - a)^3 + \cdots \]

Analyze the Sign of the Second Derivative

Next, consider the sign of \(f''(a)\). - If \(f''(a) > 0\), the leading term \(\frac{f''(a)}{2}(x - a)^2\) is positive for all \(x\) near \(a\), indicating that \(f(x) > f(a)\), suggesting a minimum at \(x = a\). - If \(f''(a) < 0\), the leading term \(\frac{f''(a)}{2}(x - a)^2\) is negative for all \(x\) near \(a\), indicating that \(f(x) < f(a)\), suggesting a maximum at \(x = a\).

Conclusion

By considering the Taylor series expansion and the conditions \(f'(a) = 0\) and the sign of \(f''(a)\), the 'second derivative test' is verified: - \(f''(a) > 0\) implies a local minimum at \(x = a\). - \(f''(a) < 0\) implies a local maximum at \(x = a\).

Key concepts

Second Derivative Test

The second derivative test helps us determine whether a function has a local maximum or minimum at a particular point. Using this test, we examine the second derivative of a function at a point where the first derivative equals zero.
Here’s the step-by-step way to use the second derivative test:
First, find the first derivative of the function, denoted as \( f'(x) \). Set \( f'(x) = 0 \) and solve for \( x \). This gives you the critical points.
Next, take the second derivative of the function, \( f''(x) \). Evaluate the second derivative at each critical point.
There are two possible outcomes:

• If \( f''(a) > 0 \), the function has a local minimum at \( x = a \).
• If \( f''(a) < 0 \), the function has a local maximum at \( x = a \).

The intuition behind this is that a positive second derivative implies the function is concave up (shaped like a smile), indicating a minimum at that point. Conversely, a negative second derivative implies the function is concave down (shaped like a frown), indicating a maximum.

Maximum and Minimum Points

Identifying maximum and minimum points of a function is crucial in mathematical analysis and real-world applications such as optimization problems. These points can provide valuable insights into the behavior and properties of functions.
First, let's define these points:

• A local maximum point is where the function reaches a highest value in its neighborhood.
• A local minimum point is where the function reaches a lowest value in its neighborhood.

To find these points, you follow a few steps:
Find the first derivative \( f'(x) \) and set it to zero to get the critical points.
Use the second derivative test to determine the nature of these critical points as either maxima or minima:

• If \( f''(a) > 0 \), it’s a local minimum.
• If \( f''(a) < 0 \), it’s a local maximum.

Sometimes, critical points can be points of inflection, where the concavity changes but there's no maximum or minimum. For holistic analysis, you should also consider the function's endpoints and other boundaries if you're working within a restricted domain.

Differentiation

Differentiation is a fundamental concept in calculus used to find the rate of change of a function. This involves computing the derivative of a function. Let's break it down:
The derivative of a function \( f(x) \), denoted by \( f'(x) \) or \( \frac{df}{dx} \), represents the instantaneous rate of change of the function at any point \( x \).
Here’s how you find the derivative:
The common rules of differentiation include:

• Power Rule: If \( f(x) = x^n \), then \( f'(x) = nx^{n-1} \).
• Product Rule: If \( f(x) = u(x)v(x) \), then \( f'(x) = u'(x)v(x) + u(x)v'(x) \).
• Quotient Rule: If \( f(x) = \frac{u(x)}{v(x)} \), then \( f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2} \).
• Chain Rule: If a function is composed of other functions, \( f(x) = g(h(x)) \), then \( f'(x) = g'(h(x)) \times h'(x) \).

Differentiation can be applied multiple times to find higher-order derivatives. For example, the second derivative \( f''(x) \) is the derivative of the first derivative \( f'(x) \). Higher-order derivatives provide information about the curvature and concavity of the function, which is used in tests like the second derivative test.