Question

Arithmetic Mean The arithmetic mean of two numbers \(a\) and \(b\) is \((a+b) / 2 .\) Show that for \(f(x)=x^{2}\) on any interval \([a, b],\) the value of \(c\) in the conclusion of the Mean Value Theorem is \(c=(a+b) / 2 .\)

Short answer

Yes, the value of 'c' in the conclusion of the Mean Value Theorem for the function \(f(x)=x^{2}\) on any interval [a, b] is indeed the arithmetic mean of 'a' and 'b', \(c=(a+b) / 2 \).

Step-by-step solution

Find derivative of \(f(x)\)

First, let's compute the derivative of the given function \(f(x)=x^{2}\). Using power rule, \(f'(x) = 2x\). This derivative will be used in the conclusion of the Mean Value Theorem.

Apply Mean Value Theorem

The Mean Value Theorem states that there exists at least one number 'c' in interval \([a, b]\) such that \(f'(c) = \frac{f(b) - f(a)}{b - a}\). Plug in our function and its derivative in this formula: For the given function \(f(x)\), we have \(f(a) = a^2\) and \(f(b) = b^2\). Now, plug these into the Mean Value Theorem formula: \(2c = \frac{b^2 - a^2}{b - a}\)

Simplify the Mean Value Theorem Equation

Now, rewrite the right hand side using difference of squares formula \(a^2-b^2 = (a+b)(a-b)\) to simplify the equation. Now, the equation becomes: \(2c = \frac{(b + a)(b - a)}{b - a}\). This simplifies to: \(2c = b+a\)

Find Solution for 'c'

Finally, solve for 'c': \(c = \frac{b + a}{2}\).

Key concepts

Arithmetic Mean

The arithmetic mean, often called the average, is a concept most students encounter early on in mathematics. It represents a value that provides a central point or 'average' for a set of numbers. In its simplest form, for two numbers, the arithmetic mean is calculated by adding the two numbers together and then dividing by two, as denoted by \( \frac{a+b}{2} \). This calculation is significant in the Mean Value Theorem, as it helps establish a specific point, referred to as 'c', that provides an average slope or rate of change over an interval on a curve. Understanding how an arithmetic mean works is crucial when exploring how functions behave over an interval, an idea central to calculus.

Derivative

In calculus, a derivative represents the rate at which a function is changing at any point. In simpler terms, it measures how a function's output value moves as its input changes. For a function \( f(x) \), the derivative is denoted by \( f'(x) \). The process of calculating a derivative is called differentiation. The derivative tells us about the slope of the function at any given point. For instance, a derivative of zero indicates that the function has reached a plateau, where it is neither increasing nor decreasing. Derivatives are foundational to calculus and are particularly vital in the Mean Value Theorem, which asserts that there is at least one point where the function's instantaneous rate of change (its derivative) matches the average rate of change over the entire interval.

Power Rule

The power rule is a quick and efficient tool in differential calculus for finding the derivative of a function where the variable has an exponent. It states that if you have a function of the form \( f(x) = x^n \), where 'n' is any real number, the derivative of that function, \( f'(x) \), is \( nx^{n-1} \). This rule simplifies the process of differentiation, especially in cases where the function is a simple polynomial. In the context of the Mean Value Theorem exercise, the power rule is applied to \( f(x) = x^2 \) to obtain the derivative \( f'(x) = 2x \), which then helps identify the specific point \( c \) that meets the conditions set by the theorem.

Difference of Squares

The difference of squares is a mathematical pattern that emerges when a squared number is subtracted from another squared number. The formula is expressed as \( a^2 - b^2 = (a + b)(a - b) \), which reveals that the difference between two squares can be factored into the product of their sum and their difference. This property comes in handy when simplifying complex algebraic expressions, as seen in the solution to the Mean Value Theorem problem we're discussing. By recognizing that \( b^2 - a^2 \) is a difference of squares, you can simplify the equation considerably, allowing you to solve for the value of \( c \) that conforms to the theorem's constraints.