Problem 28 - Essential Calculus Early Transcendentals

Question

Suppose \(f\) is an odd function and is differentiable everywhere. Prove that for every positive number \(b,\) there exists a number \(c\) in \((-b, b)\) such that \(f^{\prime}(c)=f(b) / b\)

Step-by-Step Solution

Answer

The statement is proven by applying the Mean Value Theorem using the properties of odd functions.

Step 1: Understand the Problem

We need to show that for any positive number \(b\), there exists a number \(c\) within the interval \((-b, b)\) such that \(f'(c) = \frac{f(b)}{b}\). Given that \(f\) is an odd function and differentiable everywhere, recognize that \(f(x) = -f(-x)\).

Step 2: Apply the Odd Function Property

Since \(f\) is an odd function, it follows that \(f(b) = -f(-b)\). This is based on the property \(f(x) = -f(-x)\). Hence, we have new expressions for the values at \(b\) and \(-b\) that can be used for further derivations.

Step 3: Use the Mean Value Theorem (MVT)

The Mean Value Theorem states that if a function \(g\) is continuous on \([a, b]\) and differentiable on \((a, b)\), then there exists some \(c\) in \((a, b)\) such that \(g'(c) = \frac{g(b) - g(a)}{b - a}\). Apply this to \(f(x)\) on the interval \([-b, b]\).

Step 4: Calculate Using MVT

Applying the MVT to \(f(x)\) on \([-b, b]\), we get \(f'(c) = \frac{f(b) - f(-b)}{b - (-b)} = \frac{f(b) + f(b)}{2b}\), since \(f(-b) = -f(b)\). This simplifies to \(f'(c) = \frac{2f(b)}{2b} = \frac{f(b)}{b}\).

Step 5: Conclude Existence of c

By the MVT application, we have shown that there exists some \(c\) in \((-b, b)\) where \(f'(c) = \frac{f(b)}{b}\), thus proving the statement.

Key Concepts

Odd Function

An odd function is one where the function visually mirrors around the origin of the graph. In mathematical terms, if you have an odd function \(f\), it satisfies the condition that \(f(x) = -f(-x)\). This means that if you input a positive value \(x\), the value at \(f(x)\) is the negative of the value you'd get if you input \(-x\).
This symmetry property helps in solving problems like the one provided. For example, if \(f\) is odd and differentiable, it implies that \(f(b) = -f(-b)\) for any positive number \(b\).

• Symmetrical around the origin
• Key property: \(f(x) = -f(-x)\)
• Used to simplify expressions when proving properties

This attribute is crucial when applying the Mean Value Theorem (MVT) for such functions over symmetric intervals to identify particular values like \(c\) in our problem.

Derivative

The derivative of a function gives us the rate of change or slope of the function at any point along its curve. For differentiable functions, it can be interpreted as the function's steepness at any given position \(c\).
In calculus, the derivative \(f'(x)\) is defined as the limit:\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]This formula helps to compute the change in \(f(x)\) as your input \(x\) changes infinitesimally.

• Measures slope of the tangent line
• Enables finding instantaneous rates of change
• Integral part of applying the Mean Value Theorem (MVT)

In our main question, using the derivative, we can find the value \(f'(c)\) within the interval \((-b, b)\), affirming the conditions for the derivative's existence based on MVT.

Calculus Proof

When proving properties in calculus, particularly with the Mean Value Theorem (MVT), we often start by identifying given conditions and applying known theorems to show existence or equality.
The Mean Value Theorem states:- If a function \(g\) is continuous on a closed interval \([a, b]\), and differentiable on an open interval \((a, b)\), then there's at least one point \(c\) in \((a, b)\) such that: \[ g'(c) = \frac{g(b) - g(a)}{b - a} \]This concept allows us to identify points where the function's rate of change (derivative) matches the average rate over that interval.

• Core tool for proving existence of function behavior
• Relates finite intervals with differential behavior
• Helps in showing equivalencies within specified ranges

By applying MVT to an odd, differentiable function over the interval \([-b, b]\), we find the specific \(c\) which satisfies \(f'(c) = \frac{f(b)}{b}\), thus completing the proof.