Question

Suppose that \(f\) is continuous on \([a, b], f_{+}^{\prime}(a)\) exists, and \(\mu\) is between \(f_{+}^{\prime}(a)\) and \((f(b)-f(a)) /(b-a)\). Show that \(f(c)-f(a)=\mu(c-a)\) for some \(c\) in \((a, b)\).

Short answer

Question: Find a point \(c\) in the open interval \((a, b)\) such that \(f(c) - f(a) = \mu (c - a)\), where \(f\) is a continuous function on \([a, b]\), the right-hand derivative \(f_{+}^{\prime}(a)\) exists, and a value \(\mu\) is given between \(f_{+}^{\prime}(a)\) and the average rate of change of \(f\) on the interval \([a, b]\). Answer: To find the point \(c\) in the open interval \((a, b)\) such that \(f(c) - f(a) = \mu (c - a)\), we first define a function \(g(x) = f(x) - \mu (x - a)\). We then compute \(g(a)\) and \(g(b)\), showing that they have opposite signs. Using the Intermediate Value Theorem, we can conclude that there exists a point \(c \in (a, b)\) such that \(g(c) = 0\). Finally, we can show that \(f(c) - f(a) = \mu (c - a)\) by substituting \(g(c) = 0\) back into the function definition.

Step-by-step solution

Define the function g(x)

Let's define a function \(g(x) = f(x) - \mu (x - a)\), where \(\mu\) is given between \(f_{+}^{\prime}(a)\) and \((f(b)-f(a))/(b-a)\).

Find g(a) and g(b)

Now let's compute \(g(a)\) and \(g(b)\): - \(g(a) = f(a) - \mu (a - a) = f(a)\) - \(g(b) = f(b) - \mu (b - a)\)

Show that g(a) and g(b) have opposite signs

Now, we will evaluate the given statement that \(\mu\) is between \(f_{+}^{\prime}(a)\) and \((f(b)-f(a))/(b-a)\): - Since \(\mu > f_{+}^{\prime}(a)\), we can write this as \(\mu - f_{+}^{\prime}(a) > 0\). Recall the definition of the right-hand derivative: \(\lim_{x \to a^+} \frac{f(x) - f(a)}{x - a} = f_{+}^{\prime}(a)\). Since \(\mu - f_{+}^{\prime}(a) > 0\), we conclude that \(\lim_{x \to a^+} \frac{g(x)}{x - a} > 0\). This implies that on the right side of \(x = a\), \(g(x) > 0\). - Since \(\mu < \frac{f(b) - f(a)}{b - a}\), we can write this as \((b - a) \mu - (f(b) - f(a)) < 0 \Rightarrow f(a) - f(b) + (b - a) \mu > 0\). This means \(g(b) > f(a)\). Since \(g(a) = f(a) < g(b)\) and \(g\) is continuous, by the Intermediate Value Theorem, there exists a point \(c \in (a, b)\) such that \(g(c) = 0\).

Show that \(f(c) - f(a) = \mu (c - a)\)

Since \(g(c) = 0\), we can write \(f(c) - \mu (c - a) = 0\). Hence, \(f(c) - f(a) = \mu (c - a)\). So, there exists a point \(c\) in the interval \((a, b)\) satisfying the given condition.

Key concepts

Continuity

Continuity is a fundamental concept in mathematics, especially in calculus. It describes a property of functions where they do not have any abrupt changes, jumps, or breaks.A function is said to be continuous on an interval if you can draw the graph on that interval without lifting your pen from the paper.

For a function \( f \) to be continuous on an interval \([a, b]\), it must satisfy two key conditions:

• The function \( f \) must be defined at every point in the interval \([a, b]\).
• The limit \( \lim_{x \to c} f(x) = f(c) \) must hold true for every point \( c \) within \([a, b]\).

When working with problems involving the Intermediate Value Theorem, like the one given, continuity is crucial.This theorem relies on the continuity of the function over the interval to ensure that all intermediate values are achieved.Thus, since \( f \) is continuous on \([a, b]\), we can confidently apply the Intermediate Value Theorem to find the value of \( c \) as described.

Right-hand derivative

The right-hand derivative is a specific type of derivative that focuses on the behavior of a function as it approaches a point from the right.It provides us with the rate of change of a function just as we are entering a point from greater values than the specified point.

Mathematically, the right-hand derivative of a function \( f \) at a point \( a \) is defined as:\[ f_{+}^{\prime}(a) = \lim_{x \to a^+} \frac{f(x) - f(a)}{x - a}.\]This expression looks at the limit of the difference quotient as \( x \) comes from the right. In the exercise, \( f_{+}^{\prime}(a) \) indicates the initial slope at the beginning of the interval \( a \), and its existence implies the function is smoothly transitioning as it moves into \( a \).

Understanding this concept is key to showing why \( \mu \), being between \( f_{+}^{\prime}(a) \) and the average rate of change over \([a, b]\), ensures the conditions for our Intermediate Value Theorem application are valid.This demonstrates that \( \mu \) maintains a position between the immediate slope entering \( a \) and the broader slope considered over the whole interval.

Mathematical proof

A mathematical proof is a logical argument that verifies the truth of a statement using previously established facts and rules. The purpose of a proof is to provide an indisputable reasoning for why a mathematical statement or proposition is true.

In this exercise, we construct a proof using several fundamental concepts, like continuity and the Intermediate Value Theorem. The steps involve:

• Defining a function: Introduce \( g(x) = f(x) - \mu(x - a) \), incorporating \( \mu \) directly.
• Evaluating boundary conditions: Determine \( g(a) \) and \( g(b) \) to understand behavior at the edges, observing that their signs differ.
• Using the Intermediate Value Theorem: Show a value \( c \) must exist between \( a \) and \( b \) where \( g(c) = 0 \), consequently proving \( f(c) - f(a) = \mu(c - a) \).

Each step strategically builds on definitions, properties, and theorems to confirm the truth of the main statement.Through these steps, we achieve a robust proof that elaborates not only the mathematics but also reinforces reasoning skills.This journey through logical deduction showcases how each mathematical property plays a role in developing a thorough understanding.