Problem 161 - Calculus Volume 1

Question

Use the Mean Value Theorem and find all points \(\quad 0

Step-by-Step Solution

Answer

The point \( c \) is \( \sqrt{\frac{4}{3}} \).

Step 1: Verify the Conditions of the Mean Value Theorem

The Mean Value Theorem states that if a function \( f(x) \) is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists at least one point \( c \) in \((a, b)\) where \( f'(c) = \frac{f(b) - f(a)}{b - a} \). Since \( f(x) = x^3 \) is a polynomial, it is continuous and differentiable everywhere. Therefore, the conditions of the Mean Value Theorem are satisfied on \([0, 2]\).

Step 2: Calculate \( f(2) - f(0) \)

Substitute \( x = 2 \) and \( x = 0 \) into the function \( f(x) = x^3 \) to find \( f(2) \) and \( f(0) \):- \( f(2) = 2^3 = 8 \)- \( f(0) = 0^3 = 0 \)Thus, \( f(2) - f(0) = 8 - 0 = 8 \).

Step 3: Set Up the MVT Equation

According to the Mean Value Theorem, we have:\[f'(c) = \frac{f(2) - f(0)}{2 - 0} = \frac{8}{2} = 4\]We need to find \( c \) such that \( f'(c) = 4 \).

Step 4: Find the Derivative \( f'(x) \)

Find the derivative of the function \( f(x) = x^3 \):\[ f'(x) = \frac{d}{dx}(x^3) = 3x^2 \]

Step 5: Solve for \( c \)

We need to solve the equation \( f'(c) = 4 \) for \( c \):\[ 3c^2 = 4 \]Divide both sides by 3:\[ c^2 = \frac{4}{3} \]Take the square root of both sides:\[ c = \pm \sqrt{\frac{4}{3}} \]Since \( c \) must be in the interval \((0, 2)\), we choose the positive root:\[ c = \sqrt{\frac{4}{3}} \]

Step 6: Determine Validity of \( c \)

Calculate \( \sqrt{\frac{4}{3}} \) to check if \( c \) is indeed between 0 and 2:\[ \sqrt{\frac{4}{3}} \approx 1.15 \]Since \( 1.15 \) is within the interval \( (0, 2) \), \( c = \sqrt{\frac{4}{3}} \) is a valid solution.

Key Concepts

Understanding Derivatives

Derivatives are a fundamental concept in calculus. They tell us how fast a function is changing at any given point.
For a function, the derivative can be thought of as the slope of the tangent line to the function at that point.
Calculating derivatives is key when working with the Mean Value Theorem, as it involves finding rates of change.

• The derivative of a function \( f(x) \) is denoted as \( f'(x) \).
• To find the derivative of a polynomial function, such as \( x^3 \), we use the power rule.
• The power rule states that the derivative of \( x^n \) is \( n \cdot x^{n-1} \).

For our function \( f(x) = x^3 \), applying the power rule:\[ f'(x) = 3x^2 \]This new function, \( f'(x) = 3x^2 \), gives the rate of change of \( x^3 \), or the slope at any point \( x \) on the curve of the function.
Understanding how derivatives work is crucial to finding points \( c \) that satisfy the Mean Value Theorem.

Continuous Functions: What They Are and Why They Matter

A continuous function is a type of mathematical function that has no breaks, holes, or gaps in its graph.
You can think of it like a smooth, unbroken line that you can draw without lifting your pen off the paper.
This quality of being unbroken means that a continuous function takes on every value smoothly within an interval.

• For the Mean Value Theorem to apply, the function must be continuous on the closed interval \([a, b]\).
• Polynomials, like \( f(x) = x^3 \), are continuous everywhere because they are made up of simple operations like addition, subtraction, and multiplication.
• Thus, for any polynomial function, the continuity condition required by the Mean Value Theorem is always satisfied.

In our specific exercise, because \( f(x) = x^3 \) is a continuous polynomial, the Mean Value Theorem can be used to find suitable values of \( c \) in the given interval \((0, 2)\).
This continuity essentially allows us to find at least one point within the interval where the instantaneous rate of change (derivative) equals the average rate of change over the interval.

Polynomials and Their Properties

Polynomials are algebraic expressions that involve sums of powers of a variable.
They are the most basic types of functions in mathematics and possess key features that make them ideal for many calculus problems, including those involving the Mean Value Theorem.
Polynomials can be plotted as smooth, predictable curves on a graph.

• A polynomial function of degree \( n \) is of the form: \( a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 \)
• Unlike other types of functions, polynomials are continuous and differentiable everywhere.
• The degree of the polynomial tells you the maximum number of times the curve can change direction.

In the Mean Value Theorem exercise, \( f(x) = x^3 \) is a simple polynomial function with a degree of 3.
This means it can bend up to 2 times, forming curves that change the direction once before flattening out again.
Since it is continuous and differentiable everywhere, it meets all the criteria necessary for the Mean Value Theorem, ensuring that the derivative method can be used to find specific points \( c \) in the given interval.