Question

A function \(f(x)\) and interval \([a, b]\) are given. Check if the Mean Value Theorem can be applied to \(f\) on \([a, b] ;\) if so, find a value \(c\) in \([a, b]\) guaranteed by the Mean Value Theorem. \(f(x)=x^{3}-2 x^{2}+x+1\) on [-2,2]

Short answer

The Mean Value Theorem applies, with \( c = 2 \) and \( c = -\frac{2}{3} \) in \([-2, 2]\).

Step-by-step solution

Verify Continuity on [a, b]

First, check if the function \( f(x) = x^3 - 2x^2 + x + 1 \) is continuous on the interval \([-2, 2]\). Polynomial functions are continuous everywhere, so \( f(x) \) is continuous on \([-2, 2]\).

Verify Differentiability on (a, b)

Next, check if the function is differentiable on the open interval \((-2, 2)\). Polynomial functions are differentiable everywhere, so \( f(x) \) is differentiable on \((-2, 2)\).

Check Application Conditions for Mean Value Theorem

Since the function is both continuous on \([-2, 2]\) and differentiable on \((-2, 2)\), the Mean Value Theorem (MVT) can be applied. The MVT states that there exists at least one point \( c \) in \( (a, b) \) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \).

Calculate \( f'(x) \)

To apply the MVT, first find the derivative of \( f(x) \). For \( f(x) = x^3 - 2x^2 + x + 1 \), the derivative is \( f'(x) = 3x^2 - 4x + 1 \).

Calculate \( f(a) \) and \( f(b) \)

Calculate \( f(-2) \) and \( f(2) \). Start with evaluating \( f(-2) = (-2)^3 - 2(-2)^2 + (-2) + 1 = -8 - 8 - 2 + 1 = -17 \). Then evaluate \( f(2) = 2^3 - 2(2)^2 + 2 + 1 = 8 - 8 + 2 + 1 = 3 \).

Apply the Mean Value Theorem

Calculate the slope of the line joining the points \((-2, f(-2))\) and \((2, f(2))\): \( \frac{f(2) - f(-2)}{2 - (-2)} = \frac{3 - (-17)}{4} = \frac{20}{4} = 5 \). According to MVT, set \( f'(c) = 5 \) and solve for \(c\).

Solve for \( c \)

Equate the derivative to the slope from the MVT: \( 3c^2 - 4c + 1 = 5 \). Solve by simplifying: \( 3c^2 - 4c + 1 - 5 = 0 \), resulting in \( 3c^2 - 4c - 4 = 0 \). Use the quadratic formula, \( c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3, b = -4, \) and \( c = -4 \). Calculate \( c \).

Calculate Quadratic Formula

Calculate \( c \) using the quadratic formula: \( c = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \times 3 \times (-4)}}{2 \times 3} = \frac{4 \pm \sqrt{16 + 48}}{6} = \frac{4 \pm \sqrt{64}}{6} = \frac{4 \pm 8}{6} \). This gives potential solutions \( c = 2 \) and \( c = -\frac{2}{3} \). Since \( c \) must lie in the interval \((-2, 2)\), both values are valid.

Key concepts

Understanding Polynomial Functions

Polynomial functions are mathematical expressions that consist of variables and coefficients involving terms with non-negative integer exponents. A polynomial function can be described in the general form:

• The constant terms, independent of variables.
• The terms involving variables raised to the power, such as quadratic or cubic terms.

In the original exercise, the given polynomial is: \[ f(x) = x^3 - 2x^2 + x + 1 \] This function includes a cubic term, a quadratic term, a linear term, and a constant. Polynomial functions, like these, have smooth and continuous curves, making them easy to graph and analyze over any interval.

Continuity in Polynomial Functions

Continuity in a function signifies that a small change in the input leads to a small change in the output, without any abrupt jumps or interruptions. For any polynomial function like the one in our example \[ f(x) = x^3 - 2x^2 + x + 1 \] continuity is a given.

• By nature, polynomial functions are continuous everywhere on the real line, meaning they have no breaks, holes, or gaps.
• In the interval \([-2, 2]\), the given function is continuous without exceptions.

This continuity is crucial when applying the Mean Value Theorem, confirming the function behaves smoothly over the specified interval.

Differentiability of Polynomial Functions

Differentiability is related to a function's ability to have a derivative at every point in an interval. For polynomial functions, like:\[ f(x) = x^3 - 2x^2 + x + 1 \] this property is inherent.

• A function is differentiable on an interval if it has a derivative at each point in that interval.
• Differentiability implies that the function's graph has a tangent line at each point, further implying smoothness without sharp turns.
• In the open interval \((-2, 2)\), our polynomial function is smooth and allows for easy computation of its derivative: \(f'(x) = 3x^2 - 4x + 1\).

This differentiability complements the function's continuity, making it eligible for the Mean Value Theorem.

Applying the Quadratic Formula

The quadratic formula is used to find the roots of a quadratic equation: \[ ax^2 + bx + c = 0 \] This formula is a reliable tool, particularly useful when solutions are not obvious: \[ c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] In the exercise, it was key to find values of \(c\) that satisfy the Mean Value Theorem: \[ 3c^2 - 4c - 4 = 0 \]

• This equation is derived from setting the derivative equal to the slope found.
• By substituting \(a = 3\), \(b = -4\), and \(c = -4\) into the formula, we determined that the potential solutions were \(c = 2\) and \(c = -\frac{2}{3}\).
• Each solution needs to lie within the open interval \((-2, 2)\) to be valid according to the theorem.

Both solutions fell within the interval, ensuring the Mean Value Theorem's conditions were satisfied.