Problem 65 - Essential Calculus Early Transcendentals

Question

Prove Fermat's Theorem for the case in which \(f\) has a local minimum at \(c .\)

Step-by-Step Solution

Answer

Since \( f'(c) = \lim_{h \to 0} \frac{f(c+h) - f(c)}{h} = 0 \), Fermat's theorem is verified for a local minimum.

Step 1: Understand Fermat's Theorem

Fermat's Theorem states that if a function \( f \) has a local extremum (maximum or minimum) at a point \( c \), and if the derivative \( f'(c) \) exists, then \( f'(c) = 0 \). In this exercise, we focus on the case where \( f \) has a local minimum at \( c \).

Step 2: Define a Local Minimum

A function \( f \) has a local minimum at \( c \) if there exists an interval \((a, b)\) containing \( c \) such that for every \( x \) in this interval, \( f(x) \geq f(c) \).

Step 3: Apply Derivative Definition

Since \( f \) has a local minimum at \( c \), \( c \) is a critical point where \( f'(c) \) exists. By definition, if \( f \) is differentiable at \( c \), then the derivative is given by:\[f'(c) = \lim_{h \to 0} \frac{f(c+h) - f(c)}{h}\]

Step 4: Analyze the Behavior of \( f \) Near \( c \)

For \( x = c + h \) with \( h > 0 \) and \( h \to 0 \), because \( f(c) \) is a minimum, \( f(c+h) \geq f(c) \). Therefore, the difference quotient \( \frac{f(c+h) - f(c)}{h} \geq 0 \). For \( x = c - h \) with \(-h < 0\) and \( h \to 0 \), \( f(c-h) \geq f(c) \). Thus, \( \frac{f(c-h) - f(c)}{-h} \leq 0 \).

Step 5: Calculate Limits Approach

Both limits \( \lim_{h \to 0^+} \frac{f(c+h) - f(c)}{h} \) and \( \lim_{h \to 0^-} \frac{f(c-h) - f(c)}{-h} \) must be equal for \( f'(c) \) to exist. As these are nonpositive and nonnegative respectively, they must both be zero.

Step 6: Conclude the Proof

Since the limit of the difference quotient as \( h \to 0 \) is 0, it follows that \( f'(c) = 0 \). Thus, Fermat's Theorem is proved for the case where \( f \) has a local minimum at \( c \).

Key Concepts

Local Minimum

A local minimum is a key concept in understanding Fermat's Theorem. It refers to a point in a function where the function's value is lower than all other values in its immediate vicinity. For a function \( f \) to have a local minimum at a point \( c \), there must exist an interval around \( c \), say \((a, b)\), where for every point \( x \) within this interval, the inequality \( f(x) \geq f(c) \) holds true.

This means that \( c \) is the lowest point within the interval \((a, b)\). However, it doesn't have to be the lowest value in the entire function, just locally. Understanding local minimum is crucial because it helps identify where the derivative of the function equals zero at critical points, which is the essence of Fermat's Theorem.

• Local Minimum: \( f(x) \geq f(c) \) for \( x \) in \((a, b)\)
• Not necessarily the global minimum

Differentiability

Differentiability refers to whether a function has a derivative at a specific point. For Fermat's Theorem to apply, the function must be differentiable at the point \( c \) where it has a local minimum. When a function is differentiable at a point, you can compute the derivative, which gives the slope of the tangent line to the function at that point.

If a function \( f \) is differentiable at \( c \), then the derivative is given by the limit:\[ f'(c) = \lim_{h \to 0} \frac{f(c+h) - f(c)}{h} \]

• Differentiable: Function's derivative exists at a point
• Provides slope of the tangent line

Critical Point

A critical point of a function is a point where the derivative of the function is zero or undefined. In the context of Fermat's Theorem and the exercise we are focusing on, a critical point is important because it indicates where the function might have a local minimum or maximum. This is because the slope of the tangent (represented by the derivative) is zero at these points.

When \( f'(c) = 0 \), \( c \) could be a local minimum or maximum, or even a saddle point. Determining the exact nature of the critical point may require additional analysis, but for Fermat's Theorem, recognizing \( c \) as a critical point implies that \( f \) possibly has a local extremum there.

• Critical Points: where \( f'(c) = 0 \) or undefined
• Possible local extremum

Derivative

The derivative of a function measures how the function value changes as its input changes. It represents the slope of the tangent line to the graph of the function at any point. In Fermat's Theorem, the derivative at the critical point \( c \) plays a crucial role because it determines whether \( c \) can be a point of local minimum or maximum.

When calculating the derivative at \( c \), we consider the limits as \( h \to 0 \). If the limits of the difference quotients on both the positive and negative sides equal zero, then \( f'(c) = 0 \). This indicates a flat slope at \( c \), suggesting a possible local extremum.

• Derivative: \( f'(x) = \frac{df}{dx} \)
• Represents rate of change

Mathematical Proof

A mathematical proof is a logical argument demonstrating the truth or falsehood of a given proposition, based on deductive reasoning. In proving Fermat's Theorem for a local minimum, we systematically showed that if \( f \) has a local minimum at \( c \) and is differentiable, then \( f'(c) = 0 \).

The proof involves:

• Defining a local minimum
• Establishing the derivative's existence
• Analyzing how changes around \( c \) affirm that \( f'(c) = 0 \)
• Concluding with logical certainty

The steps followed in the proof provide a clear, logical flow of reasoning to ensure that the theorem holds true under its specified conditions.