Problem 2 - 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

All conditions are satisfied. \( c = \frac{2 + \sqrt{7}}{3} \) satisfies Rolle's theorem.

Step 1: Verify Continuity

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

Step 2: Verify Differentiability

Check if the function \( f(x) = x^3 - 2x^2 - x + 2 \) is differentiable on the open interval \((1, 2)\). Polynomials are differentiable everywhere, so \( f(x) \) is differentiable on \((1, 2)\).

Step 3: Verify Function Values at Endpoints

Check if \( f(1) = f(2) \). Calculate: \( f(1) = 1^3 - 2(1)^2 - 1 + 2 = 0 \) and \( f(2) = 2^3 - 2(2)^2 - 2 + 2 = 0 \). Both are equal, so \( f(1) = f(2) \).

Step 4: Apply Rolle's Theorem

Since all conditions of Rolle's theorem are satisfied, there must exist a \( c \in (1, 2) \) such that \( f'(c) = 0 \). First, find the derivative: \( f'(x) = 3x^2 - 4x - 1 \).

Step 5: Solve for \( c \)

Set \( f'(c) = 0 \) and solve for \( c \): \( 3c^2 - 4c - 1 = 0 \). Use the quadratic formula \( c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ c = \frac{4 \pm \sqrt{16 + 12}}{6} = \frac{4 \pm \sqrt{28}}{6} = \frac{4 \pm 2\sqrt{7}}{6} \]. Simplify to \( c = \frac{2 \pm \sqrt{7}}{3} \). Since \( c \) must be in \((1, 2)\), select \( c = \frac{2 + \sqrt{7}}{3} \).

Key Concepts

Continuity of Polynomial Functions

Polynomials are some of the most familiar and straightforward functions in mathematics. A polynomial function is a function that is expressed as a sum of terms, each consisting of a variable raised to a non-negative integer power, multiplied by a coefficient. For example, the function given in the exercise, \[ f(x) = x^3 - 2x^2 - x + 2 \], is a polynomial of degree 3. One of the essential properties of polynomials is their continuity. This means that polynomial functions do not have breaks, jumps, or holes in their graphs. They are *continuous functions* everywhere on the real number line. Hence, a polynomial function is *continuous* in any interval you consider, whether it's an open or closed interval. In our exercise, it was necessary to check if \( f(x) \) is continuous on the closed interval \([1, 2]\). Since all polynomial functions are inherently continuous, this condition is automatically satisfied, making our task straightforward in verifying Rolle's theorem.

Differentiability of Polynomial Functions

Just like continuity, differentiability is another crucial property of polynomial functions. A function is said to be differentiable at a point if its derivative exists at that point, and it is differentiable over an interval if it is differentiable at every point within that interval. For a polynomial function, this means that the rate of change of the function is smooth and unbroken, without any sharp corners or edges. In simpler terms, you should be able to draw the graph of a polynomial function without lifting your pen, and its slope will change smoothly. In our exercise, we needed to verify if \( f(x) \) is differentiable over the interval \((1, 2)\). Given that any polynomial function is differentiable everywhere on the real number line, our function \( f(x) = x^3 - 2x^2 - x + 2 \) is differentiable on the open interval \((1, 2)\). This is another condition satisfied for Rolle's theorem.

Quadratic Formula

The quadratic formula is a fundamental tool in algebra used to solve quadratic equations of the form \( ax^2 + bx + c = 0 \). It provides the solutions, or *roots*, of the equation through the formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. To apply this formula, you need to identify the coefficients \(a\), \(b\), and \(c\) from the quadratic equation. In our problem, to find the critical points where the derivative of \(f(x)\) equals zero, we had to solve the quadratic equation: \[ 3c^2 - 4c - 1 = 0 \]. Here, \(a = 3\), \(b = -4\), and \(c = -1\). Plugging these values into the quadratic formula, we obtained: \[ c = \frac{4 \pm \sqrt{16 + 12}}{6} = \frac{4 \pm \sqrt{28}}{6} = \frac{4 \pm 2\sqrt{7}}{6} \]. Simplifying further, we got: \[ c = \frac{2 \pm \sqrt{7}}{3} \]. Given the interval \( (1, 2) \), we selected the appropriate root \( c = \frac{2 + \sqrt{7}}{3} \) that lies within this range, satisfying Rolle's theorem.