Problem 165 - Calculus Volume 1

Question

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

Step-by-Step Solution

Answer

The point is \(c = 1\).

Step 1: Verify Conditions for Mean Value Theorem

The Mean Value Theorem (MVT) applies if the function \(f(x)\) is continuous on the closed interval \([0, 2]\) and differentiable on the open interval \((0, 2)\). Since \(f(x) = (x-1)^{10}\) is a polynomial, it is both continuous and differentiable everywhere, including the intervals specified. Thus, MVT can be applied.

Step 2: Apply the Mean Value Theorem Formula

According to the Mean Value Theorem, there exists some \(c\) in the interval \((0, 2)\) such that \[ f'(c) = \frac{f(2) - f(0)}{2 - 0}. \]

Step 3: Calculate \(f(2)\) and \(f(0)\)

Substitute the values into \(f(x) = (x-1)^{10}\):\[ f(2) = (2-1)^{10} = 1^{10} = 1 \]\[ f(0) = (0-1)^{10} = (-1)^{10} = 1 \]

Step 4: Compute \(f(2) - f(0)\)

Calculate the difference:\[ f(2) - f(0) = 1 - 1 = 0. \]

Step 5: Calculate the Derivative \(f'(x)\)

Find the derivative of \(f(x) = (x-1)^{10}\);\[ f'(x) = 10(x-1)^9. \]

Step 6: Solve for \(c\)

Set up the equation from the Mean Value Theorem:\[ 10(c-1)^9 = 0. \]Divide by 10:\[ (c-1)^9 = 0. \]This implies \(c-1 = 0\) or \(c = 1\).

Step 7: Verify \(c\) is in the Interval

Ensure that \(c = 1\) lies within the open interval \((0, 2)\). Since \(c = 1\) does indeed fall within this interval, it is a valid solution.

Key Concepts

Polynomials

Polynomials are equations that involve terms made up of variables raised to whole number powers and have coefficients. The polynomial used in this example is \( f(x) = (x - 1)^{10} \). This is a simple example of a polynomial, where the power is 10, and the polynomial has one variable term, \( (x-1) \).
Polynomials have several key features that make them important in calculus and the Mean Value Theorem application:

• Degree: The degree of a polynomial is the highest power of the variable in the expression. Here, the degree is 10.
• Continuity: Since every polynomial function is continuous everywhere on the real line, there are no breaks, jumps, or holes in the graph of the polynomial function.
• Roots: For any polynomial \( f(x) \), if you set \( f(x) = 0 \), the solutions are called roots. Here, if you solved \( (x-1)^{10} = 0 \), the root would be \( x = 1 \).

Polynomials like \( (x-1)^{10} \) are used in many mathematical analyses because they have predictable behaviors. When applying the Mean Value Theorem, polynomials ensure all necessary conditions are satisfied, like continuity and differentiability, which we will explore next.

Derivative

The derivative tells us how a function changes as its input changes – essentially, it's a mathematical way of describing the rate of change.
For the polynomial \( f(x) = (x-1)^{10} \), the derivative is found using the power rule:\[ f'(x) = 10(x-1)^9 \]This means, for small changes in \( x \), the value of \( f(x) \) will change at a rate proportional to the expression \( 10(x-1)^9 \). Here:

• The coefficient 10 comes from the power of the exponent in the original equation.
• The expression \((x-1)^9\) shows that the rate of change is modified by the variable \( x \), which has been reduced by 1.

Finding the derivative is essential when using the Mean Value Theorem because it shows how fast or slow the function's output is changing at any point \( c \) within the interval.

Continuity

Continuity in a function means that the graph of the function is unbroken – you can draw it without lifting your pen from the paper.
In calculus, a function being continuous over a specified interval is crucial for applying the Mean Value Theorem. With polynomials, continuity is naturally achieved because they possess no breaks, jumps, or asymptotes anywhere along their graph.
When considering the function \( f(x) = (x-1)^{10} \), we know it is continuous on any interval, including \([0, 2]\) and differentiable on \((0, 2)\).
To apply the Mean Value Theorem:

• Verify the function is continuous on the closed interval \([0, 2]\).
• Check that it is differentiable on the open interval \((0, 2)\).

Since polynomials meet these conditions, MVT ensures there exists a point \( c \) where the instantaneous rate of change (the derivative) is equal to the average rate of change over the interval.