Problem 164 - Calculus Volume 1

Question

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

Step-by-Step Solution

Answer

The point \(c = 1\) satisfies the Mean Value Theorem for this function on the interval \((0, 2)\).

Step 1: Understand the Mean Value Theorem

The Mean Value Theorem states that for a function \( f \) continuous on \([a, b]\) and differentiable on \((a, b)\), there exists a point \(c\) in \((a, b)\) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \). Here, \(a = 0\) and \(b = 2\).

Step 2: Find the Values of \(f(2)\) and \(f(0)\)

Substitute the values into the function: \( f(0) = 1 + 0 + 0^2 = 1 \) and \( f(2) = 1 + 2 + 2^2 = 7 \).

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

Compute the difference: \( f(2) - f(0) = 7 - 1 = 6 \).

Step 4: Calculate \(f'(x)\)

Differentiate \( f(x) = 1 + x + x^2 \) to find \( f'(x) = 1 + 2x \).

Step 5: Solve for \(c\) using the Mean Value Theorem

Set \( f'(c) = \frac{f(2) - f(0)}{2 - 0} = 3 \). Thus, \( 1 + 2c = 3 \). Solve for \(c\): \( 2c = 2 \), so \( c = 1 \).

Step 6: Verify the Condition \(0 < c < 2\)

Verify that \( c = 1 \) is within the interval \((0, 2)\). It satisfies \(0 < 1 < 2\).

Key Concepts

Differentiable Functions

To understand the exercise, we must first familiarize ourselves with differentiable functions. A differentiable function is one that we can differentiate, meaning it has a derivative at every point in its domain. Differentiability implies that the function is smooth and has no sharp corners or discontinuities. In the problem, the function given is \( f(x) = 1 + x + x^2 \). This quadratic function is both continuous and differentiable everywhere because it is composed of polynomial terms.

• Smoothness: No abrupt changes in behavior.
• Existence of Derivatives: You can compute the derivative \( f'(x) \) at every point in its domain. For example, the derivative of \( f(x) \) is \( f'(x) = 1 + 2x \).
• Local Linearity: On a very small scale, a differentiable function can be approximated by a linear function.

Understanding differentiable functions allows us to apply concepts such as the Mean Value Theorem effectively.

Calculus Problems

In calculus, problems like this one often revolve around finding specific values that satisfy certain conditions. The Mean Value Theorem is a powerful tool in such scenarios. The theorem states that for any function that is continuous over a closed interval \([a, b]\) and differentiable over the open interval \((a, b)\), there exists at least one point \(c\) within \((a, b)\) where the instantaneous rate of change (derivative) equals the average rate of change across the interval. For this function, those values are found by computing the average rate of change from the endpoints of the interval.With our exercise:

• We determined \( f(0) = 1 \) and \( f(2) = 7 \).
• We calculated the difference: \( f(2) - f(0) = 6 \).
• This calculation directly leads to solving for \( c \) using the derivative.

By using calculus in this manner, we can pinpoint where a function's behavior matches average change rates across intervals.

Solving Equations Step-by-Step

Solving calculus problems often involves a methodical, step-by-step approach to ensure accuracy. Let's break down the solution to our Mean Value Theorem problem in simple steps:Step 1: Verify the ConditionsEnsure that the function is continuous and differentiable over the chosen interval. Our function, \( f(x) = 1 + x + x^2 \), meets these conditions, allowing us to apply the theorem.Step 2: Differentiate the FunctionFind the derivative. For \( f(x) \), the derivative is \( f'(x) = 1 + 2x \). This step is critical as you'll use it to find \( c \).Step 3: Compute the Average Rate of ChangeCalculate \( \frac{f(2) - f(0)}{2-0} = 3 \). This value is what our derivative should equal at some point \( c \).Step 4: Set Up the Equation and Solve for \( c \)Use \( f'(c) = 3 \). This means solving \( 1 + 2c = 3 \), leading to \( c = 1 \).Step 5: Confirm \( c \) Lies Within the IntervalFinally, verify that \( c \) is indeed in the interval \((0, 2)\). Here, \( c=1 \) fits perfectly.Following these steps consistently will help you tackle similar calculus problems with confidence.