Problem 6 - The Calculus with Analytic Geometry

Question

Verify that the hypothesis of the mean-value theorem is satisfied for the given function on the indicated interval. Then find a suitable value for \(c\) that satisfies the conclusion of the mean-value theorem. $$ f(x)=x^{3}+x^{2}-x ;[-2,1] $$

Step-by-Step Solution

Answer

The value of \(c\) is \( \frac{-1 + \sqrt{7}}{3} \).

Step 1 - Verify Continuity

Ensure the function is continuous on the closed interval \([-2, 1]\). The given function \(f(x)=x^{3}+x^{2}-x\) is a polynomial, which is continuous for all real numbers. Thus, it is continuous on [-2, 1].

Step 2 - Verify Differentiability

Check that the function is differentiable on the open interval \( (-2, 1) \). Again, since \(f(x)\) is a polynomial, it is differentiable for all real numbers. Thus, it is differentiable on \( (-2, 1) \).

Step 3 - Apply Mean Value Theorem

The Mean Value Theorem (MVT) states that if \(f\) is continuous on \([a, b]\) and differentiable on \((a, b)\), then there exists at least one \(c \) in \((a, b)\) such that \(\frac{f(b) - f(a)}{b - a} = f'(c)\).

Step 4 - Compute \(f(a)\) and \(f(b)\)

Calculate \(f(-2)\) and \(f(1)\):\[f(-2) = (-2)^3 + (-2)^2 - (-2) = -8 + 4 + 2 = -2\]\[f(1) = 1^3 + 1^2 - 1 = 1 + 1 - 1 = 1\]

Step 5 - Compute \(\frac{f(b) - f(a)}{b - a}\)

Substitute \(a = -2\) and \(b = 1\) into the MVT formula:\[\frac{f(1) - f(-2)}{1 - (-2)} = \frac{1 - (-2)}{1 + 2} = \frac{1 + 2}{3} = 1\]

Step 6 - Differentiate the Function

Find the derivative \(f'(x)\):\[f'(x) = 3x^2 + 2x - 1\]

Step 7 - Solve \(f'(c) = 1\)

Set the derivative equal to the value from MVT and solve for \(c\):\[3c^2 + 2c - 1 = 1\]\[3c^2 + 2c - 2 = 0\]Solve the quadratic equation using the quadratic formula: \(c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 3\), \(b = 2\), and \(c=-2\)\[c = \frac{-2 \pm \sqrt{2^2 - 4 \times 3 \times (-2)}}{2 \times 3} = \frac{-2 \pm \sqrt{4 + 24}}{6} = \frac{-2 \pm \sqrt{28}}{6}\]\[c = \frac{-2 \pm 2\sqrt{7}}{6} = \frac{-1 \pm \sqrt{7}}{3}\]Since \( \frac{-1 + \sqrt{7}}{3} \) is in the interval \((-2, 1)\), this is our solution.

Key Concepts

Continuity

When talking about the Mean Value Theorem, continuity is a crucial concept. A function is called continuous if you can draw it without lifting your pen from the paper. In mathematical terms, it means that there are no breaks, holes, or jumps in the function. For polynomial functions, like the one given in the exercise, continuity is guaranteed for all real numbers. Specifically, for our function \(f(x) = x^3 + x^2 - x\), it is continuous on the closed interval \([-2, 1]\). This is because polynomial functions are continuous everywhere. Thus, the first condition of the Mean Value Theorem is satisfied.

Differentiability

Differentiability is the next requirement for applying the Mean Value Theorem. A function is differentiable if it has a derivative everywhere in the interval in question. The derivative of a function gives us the slope of the tangent line at any point. Again, polynomial functions like ours \(f(x) = x^3 + x^2 - x\) are differentiable everywhere because they are smooth and have no sharp corners or cusps. Thus, our function is differentiable on the open interval \((-2, 1)\), meeting the second requirement of the Mean Value Theorem.

Polynomial Functions

Polynomial functions are expressions that involve variables raised to whole number powers. These functions form curves that are smooth and unbroken. The general form of a polynomial function is \(a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0\), where \(a_n, a_{n-1}, ..., a_1, a_0\) are constants, and \(n\) is a non-negative integer. In our exercise, the function \(f(x) = x^3 + x^2 - x\) is a cubic polynomial, which means it is of the form \(ax^3 + bx^2 + cx + d\). These types of functions are always continuous and differentiable, which is why they are useful when applying the Mean Value Theorem.

Quadratic Equations

A quadratic equation is a polynomial equation of degree two, generally written as \(ax^2 + bx + c = 0\). Quadratic equations are solved using various methods like factoring, completing the square, or the quadratic formula. In our step-by-step solution, we reached a quadratic equation \(3c^2 + 2c - 2 = 0\) while applying the Mean Value Theorem. We solved this equation using the quadratic formula \(c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = 2\), and \(c = -2\). Solving the quadratic equation gave us the critical points \(c = \frac{-1 + \sqrt{7}}{3}\) and \(c = \frac{-1 - \sqrt{7}}{3}\). Only \(\frac{-1 + \sqrt{7}}{3}\) falls within our interval \((-2, 1)\), making it our solution.