Problem 2 - Calculus Early Transcendentals

Question

Verify that the hypotheses of Rolle's Theorem are satisfied on the given interval, and find all values of \(c\) in that interval that satisfy the conclusion of the theorem. $$ f(x)=x^{3}-3 x^{2}+2 x ;[0,2] $$

Step-by-Step Solution

Answer

Values of \(c\) that satisfy Rolle's Theorem are \(x = 1 + \frac{\sqrt{3}}{3}\) and \(x = 1 - \frac{\sqrt{3}}{3}\).

Step 1: Verify Continuity

Rolle's Theorem requires that the function is continuous on the closed interval \([0, 2]\). Since \(f(x) = x^3 - 3x^2 + 2x\) is a polynomial, it is continuous everywhere, including on \([0, 2]\).

Step 2: Verify Differentiability

Rolle's Theorem also requires that the function is differentiable on the open interval \((0, 2)\). Again, since \(f(x)\) is a polynomial, it is differentiable everywhere, thus it is differentiable on \((0, 2)\).

Step 3: Check Equal Function Values at Endpoints

Rolle's Theorem requires \(f(a) = f(b)\) where \(a = 0\) and \(b = 2\). Calculate \(f(0)\) and \(f(2)\). \[ f(0) = 0^3 - 3(0)^2 + 2(0) = 0 \]\[ f(2) = 2^3 - 3(2)^2 + 2(2) = 8 - 12 + 4 = 0 \]Since \(f(0) = f(2) = 0\), the condition is met.

Step 4: Solve for Critical Points

Rolle's Theorem states that there must be at least one \(c\) in \((0, 2)\) such that \(f'(c) = 0\). First, find the derivative \(f'(x)\):\[ f'(x) = 3x^2 - 6x + 2 \]Set \(f'(x) = 0\) to find \(c\):\[ 3x^2 - 6x + 2 = 0 \]

Step 5: Solve Quadratic Equation

To find the values of \(c\), solve the quadratic equation \(3x^2 - 6x + 2 = 0\) using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)where \(a = 3\), \(b = -6\), and \(c = 2\).\[ x = \frac{6 \pm \sqrt{(-6)^2 - 4(3)(2)}}{2(3)} = \frac{6 \pm \sqrt{36 - 24}}{6} = \frac{6 \pm \sqrt{12}}{6} \]Simplifying, we get: \[ x = \frac{6 \pm 2\sqrt{3}}{6} = \frac{3 \pm \sqrt{3}}{3} \]These values are \(x = 1 + \frac{\sqrt{3}}{3}\) and \(x = 1 - \frac{\sqrt{3}}{3}\).

Step 6: Validate Values in Interval

Check if these values are within the interval \((0, 2)\). Both \(1 + \frac{\sqrt{3}}{3}\) and \(1 - \frac{\sqrt{3}}{3}\) are between 0 and 2, thus both satisfy the conclusion of Rolle's Theorem in the interval \((0, 2)\).

Key Concepts

Differentiability

A function is differentiable if it has a derivative at every point in an open interval. Differentiability is essential for applying Rolle's Theorem. In simpler terms, for a graph of a function to be smooth and have no sharp turns or cusps, the function must be differentiable.
Polynomials, like the one in our exercise, are differentiable everywhere. This means they can be smoothly drawn without lifting the pencil. For our function, which is a cubic polynomial of the form \(f(x) = x^3 - 3x^2 + 2x\), it is differentiable on the entire real line.

• This implies there's no disruption or sharp point as you move across the interval \((0, 2)\).
• Because it's differentiable in this open interval, one criterion of Rolle's Theorem is satisfied.

Remember, every differentiable function is also continuous, but not every continuous function is differentiable.

Continuity

Continuity ensures that a function can be drawn without any breaks or holes over a specified interval. For Rolle’s Theorem, a continuous function is required on a closed interval \([a, b]\). In our exercise, the function \(f(x) = x^3 - 3x^2 + 2x\) is a polynomial, and polynomials are naturally continuous everywhere.

• This means from point \(a = 0\) to \(b = 2\), the graph of the function doesn't jump or break.
• It smoothly carries through, another key part of meeting the theorem’s conditions.

When a function is continuous over an interval, it indicates there are no abrupt changes in value, making it predictable and smooth across that interval.

Polynomials

Polynomials are mathematical expressions involving sums of powers of variables, and they play a crucial role in the application of Rolle’s Theorem. The function given in the exercise is a cubic polynomial.
Polynomials can be classified by the degree of their highest power. A polynomial like \(x^3 - 3x^2 + 2x\) is a degree three polynomial, or cubic, because the highest exponent is 3.

• These expressions are simple to differentiate and integrate, essential operations in calculus.
• Polynomials are inherently continuous and differentiable at every real number, simplifying many analyses.

Understanding the properties of polynomials helps in recognizing why they often satisfy the conditions necessary for Rolle’s Theorem.

Critical Points

Critical points occur where the derivative of a function is zero or undefined. In the context of Rolle’s Theorem, identifying critical points helps find where the slope of the tangent to the function is zero within the open interval \((a, b)\).
In our exercise, the derivative of the function \(f(x)=x^3-3x^2+2x\) was calculated to be \(f'(x) = 3x^2 - 6x + 2\).

• Setting \(f'(x) = 0\) yields the critical points: solutions to the equation represent where the graph is flat between endpoints.
• For this function, solving \(3x^2 - 6x + 2 = 0\) gives two critical points within \((0, 2)\).

The critical points \(x = 1 + \frac{\sqrt{3}}{3}\) and \(x = 1 - \frac{\sqrt{3}}{3}\) meet all conditions set by Rolle's Theorem for a flat tangent line within the specified interval.