Problem 176 - Calculus Volume 1

Question

Determine whether the Mean Value Theorem applies for the functions over the given interval \([a, b] .\) Justify your answer. $$ y=x^{3}+2 x+1 \text { over }[0,6] $$

Step-by-Step Solution

Answer

The Mean Value Theorem applies because the function is continuous on \([0, 6]\) and differentiable on \((0, 6)\).

Step 1: Verify Continuity

The function given is \( y = x^3 + 2x + 1 \). This is a polynomial function, and polynomial functions are continuous everywhere on the real number line. Therefore, the function is continuous on the closed interval \([0, 6]\).

Step 2: Confirm Differentiability

A polynomial function is differentiable everywhere on the real number line. Hence, \( y = x^3 + 2x + 1 \) is differentiable on the open interval \((0, 6)\).

Step 3: Apply the Mean Value Theorem Conditions

For the Mean Value Theorem (MVT) to apply, a function must be continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\). Since \( y = x^3 + 2x + 1 \) meets both conditions on the interval \([0, 6]\), the Mean Value Theorem applies.

Key Concepts

Continuity

Continuity is a fundamental concept in calculus and is crucial when discussing the Mean Value Theorem. A function is continuous on an interval if there are no breaks, jumps, or holes throughout that interval. It's like drawing a curve without lifting your pencil from the paper.
For the problem at hand, the function is a polynomial, specifically \( y = x^3 + 2x + 1 \), and it is continuous for all real numbers.

• Polynomials, such as this cubic function, have terms that include powers of the variable \(x\).
• They exhibit smooth graphs without any abrupt changes, ensuring continuity everywhere on the real number line.
• This includes our interval of interest, \([0, 6]\).

The Mean Value Theorem requires a function to be continuous on the closed interval \([a,b]\) to apply. Since our function meets this condition, it passes the continuity test for the theorem.

Differentiability

Differentiability is another key requirement for the Mean Value Theorem to hold.
It refers to the ability of a function to have a derivative at each point within an interval, indicating that the function's slope can be determined everywhere.The function \( y = x^3 + 2x + 1 \) is differentiable over the open interval \((0, 6)\). This is because:

• Polynomial functions are smooth and their derivatives exist everywhere on the real number line.
• To find a derivative, we apply the rules for differentiation to each term, resulting in \( y' = 3x^2 + 2 \), confirming differentiability.

A function being differentiable on an open interval \((a, b)\) means that there are no sharp changes in slope or points where the slope is not defined. Thus, our function is indeed differentiable on \((0, 6)\), satisfying this requirement of the Mean Value Theorem.

Polynomial Functions

Polynomial functions are a core part of algebra and calculus due to their simple yet powerful structure.
They are made up of variables raised to whole number powers, such as \( x^3 \), combined with coefficients and constant terms.
For example, the function \( y = x^3 + 2x + 1 \) is a cubic polynomial.

• It has a highest power of three, which affects the shape and behavior of its graph.
• Polynomials can have various shapes, including parabolas or "S" curves, depending on their degree.
• They are known as "nice" functions because they are both continuous and differentiable everywhere.

These properties make polynomial functions ideal candidates for applying the Mean Value Theorem, as they naturally fulfill the necessary conditions of continuity and differentiability across any specified intervals.