Problem 3 - The Calculus with Analytic Geometry

Question

Verify that conditions (i), (ii), and (iii) of the hypothesis of Rolle's theorem are satisfied by the, given function on the indicated interval. Then find a suitable value for \(c\) that satisfies the conclusion of Rolle's theorem $$ f(x)=x^{3}-2 x^{2}-x+2 ;[-1,2] $$

Step-by-Step Solution

Answer

\(c = \frac{2 + \sqrt{7}}{3}\) or \(c = \frac{2 - \sqrt{7}}{3}\)

Step 1: Verify Continuity

Check if the function \( f(x) = x^3 - 2x^2 - x + 2 \) is continuous on the interval \([-1, 2] \). Since \( f(x) \) is a polynomial, it is continuous everywhere. Thus, it is continuous on \([-1, 2] \).

Step 2: Verify Differentiability

Ensure that \( f(x) \) is differentiable on the open interval \((-1, 2)\). Since \( f(x) \) is a polynomial, it is differentiable everywhere. Thus, it is differentiable on \((-1, 2)\).

Step 3: Verify Endpoint Equality

Check that \( f(x) \) has equal values at the endpoints of the interval \([-1, 2] \). Calculate \( f(-1) \) and \( f(2) \): \[ f(-1) = (-1)^3 - 2(-1)^2 - (-1) + 2 = -1 - 2 + 1 + 2 = 0 \] \[ f(2) = (2)^3 - 2(2)^2 - 2 + 2 = 8 - 8 - 2 + 2 = 0 \] Since \( f(-1) = f(2) = 0 \), the endpoint equality condition is satisfied.

Step 4: Apply Rolle's Theorem

With all conditions of Rolle's theorem satisfied, there must be at least one \( c \) in \((-1, 2)\) such that \( f'(c) = 0 \). First, find the derivative of \( f(x) \): \[ f'(x) = 3x^2 - 4x - 1 \] Now, set \( f'(x) = 0 \): \[ 3x^2 - 4x - 1 = 0 \] Solve this quadratic equation: \( x = \frac{4 \pm\sqrt{16 + 12}}{6} = \frac{4 \pm\sqrt{28}}{6} = \frac{4 \pm 2\sqrt{7}}{6} = \frac{2 \pm\sqrt{7}}{3} \) \[ x = \frac{2 + \sqrt{7}}{3} \text{ or } x = \frac{2 - \sqrt{7}}{3} \] Both values are within the interval \((-1, 2)\).

Key Concepts

Continuity

Continuity of a function is a fundamental concept in calculus. When we say a function is continuous on an interval, it means that the function does not have any breaks, holes, or jumps within that interval. For the exercise, we have the polynomial function \( f(x) = x^3 - 2x^2 - x + 2 \).
Polynomial functions are always continuous across their entire domain, which includes all real numbers. This means there's no point within the interval \([-1, 2]\) where the function suddenly jumps to another value or has a hole. Thus, \( f(x) \) is continuous on \([-1, 2]\). This satisfies the first condition of Rolle's Theorem.

Differentiability

Differentiability is another key concept in calculus. A function is differentiable at a point if it has a well-defined tangent (slope) at that point, which means the derivative exists there. For a function to be differentiable on an interval, it must be differentiable at every point within that interval.
Our function \( f(x) = x^3 - 2x^2 - x + 2 \) is a polynomial function. Polynomial functions are differentiable everywhere on their domain, which includes all real numbers. Thus, \( f(x) \) is differentiable on the open interval \((-1, 2)\). This checks off the second condition of Rolle's Theorem.

Polynomial Functions

Polynomial functions are expressions that involve variables raised to whole number exponents and have coefficients. The general form of a polynomial function is:
\[ f(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0 \]
In our exercise, the function is:
\[ f(x) = x^3 - 2x^2 - x + 2 \]
Polynomials are essential in calculus because they are smooth and continuous, meaning they have no breaks, jumps, or sharp corners. This makes them easy to differentiate and integrate. Knowing this helps us understand why polynomial functions automatically satisfy the continuity and differentiability conditions of Rolle's Theorem.

Quadratic Equations

Quadratic equations take the form: \[ ax^2 + bx + c = 0 \].
To solve a quadratic, we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \].
In our problem, to find the values where the derivative \( f'(x) \) equals zero, we solve the quadratic:
\[ 3x^2 - 4x - 1 = 0 \]. By applying the quadratic formula:
\[ x = \frac{4 \pm \sqrt{16 + 12}}{6} = \frac{2 \pm \sqrt{7}}{3} \].
This gives us two solutions, \( x = \frac{2 + \sqrt{7}}{3} \) and \( x = \frac{2 - \sqrt{7}}{3} \). Both values lie within the interval \((-1,2)\) and satisfy the conclusion of Rolle's Theorem.