Problem 37 - The Calculus with Analytic Geometry

Question

Use Rolle's theorem to prove that the equation \(4 x^{5}+3 x^{3}+3 x-2=0\) has exactly one root that lies in the interval \((0,1)\). (HINT: First show there is at least one number in \((0,1)\) that is a root of the equation. Then assume that there is more than one root of the equation in \((0,1)\) and show that this leads to a contradiction.)

Step-by-Step Solution

Answer

The equation has exactly one root in the interval (0,1) by Rolle's Theorem and the Intermediate Value Theorem.

Step 1 - Verify the Endpoints

Check the values of the function at the endpoints of the interval: At x = 0: \[f(0) = 4(0)^5 + 3(0)^3 + 3(0) - 2 = -2\] At x = 1: \[f(1) = 4(1)^5 + 3(1)^3 + 3(1) - 2 = 4 + 3 + 3 - 2 = 8\] Since \(f(0) < 0\) and \(f(1) > 0\), by the Intermediate Value Theorem, there is at least one root in the interval \((0, 1)\).

Step 2 - Assume More Than One Root

Assume there are two roots, say \(c_1\) and \(c_2\), in the interval \((0, 1)\) where \(f(c_1) = f(c_2) = 0\).

Step 3 - Apply Rolle's Theorem

By Rolle's Theorem, if \(f(x)\) is continuous on \([c_1, c_2]\) and differentiable on \((c_1, c_2)\), there exists a point \(c\) in \((c_1, c_2)\) such that \(f'(c) = 0\).

Step 4 - Differentiate the Function

Find the derivative of the function: \[f(x) = 4x^5 + 3x^3 + 3x - 2\] \[f'(x) = 20x^4 + 9x^2 + 3\]

Step 5 - Analyze the Derivative

The derivative \(f'(x) = 20x^4 + 9x^2 + 3\) is always positive for all \(x\) in \((0,1)\) because each term is positive in this interval. Hence, \(f'(x) > 0\).

Step 6 - Derive the Contradiction

Since \(f'(x) > 0\) for all \(x\) in \((0,1)\), it contradicts the requirement of Rolle's Theorem that \(f'\) should be zero for some \(c\) in \((c_1, c_2)\). Thus, there cannot be more than one root in \((0, 1)\).

Step 7 - Conclusion

Since there is at least one root and there cannot be more than one root, the equation has exactly one root in the interval \((0, 1)\).

Key Concepts

Intermediate Value Theorem

The Intermediate Value Theorem is a fundamental result in calculus. It states that if a continuous function, say \(f\), takes values \(f(a)\) and \(f(b)\) at two points \(a\) and \(b\) respectively, then it must take any value between \(f(a)\) and \(f(b)\) at some point within \((a, b)\). In simple terms, if a function moves from a negative value to a positive value over an interval, it must cross zero at least once within that interval.
In the given problem, we checked the values at the endpoints of the interval \((0,1)\):
- At \(x = 0\): \(f(0) = -2\)
- At \(x = 1\): \(f(1) = 8\)
Since \(f(0) < 0\) and \(f(1) > 0\), by the Intermediate Value Theorem, there is at least one root of the equation \(4x^5 + 3x^3 + 3x - 2 = 0\) within the interval \((0, 1)\).

Derivative

A derivative represents the rate at which a function is changing at any given point. It gives the slope of the function's graph. The function provided in the exercise is \(f(x) = 4x^5 + 3x^3 + 3x - 2\).
To find its derivative, we use the power rule for each term:
- \((4x^5)' = 20x^4\)
- \((3x^3)' = 9x^2\)
- \((3x)' = 3\)
- The constant -2 has a derivative of 0.
This results in the derivative function \(f'(x) = 20x^4 + 9x^2 + 3\).
The derivative helps us understand the function's behavior, especially when applying Rolle's Theorem.

Root of the Equation

A root of an equation is a value of \(x\) that makes the equation equal to zero. For the equation \(4x^5 + 3x^3 + 3x - 2 = 0\), we are looking for an \(x\) in the interval \((0, 1)\) such that \(f(x) = 0\).
The Intermediate Value Theorem ensured the existence of at least one root in \((0, 1)\), as detailed in the first section. We initially assumed there might be more than one root, but then used calculus concepts to prove otherwise.

Continuity

A function is continuous if it has no breaks, gaps, or jumps in its graph. This means you can draw it without lifting your pen from the paper. For Rolle's Theorem to be applicable, the function must be continuous on the closed interval \[a, b\].
In our example, \(f(x) = 4x^5 + 3x^3 + 3x - 2\) is a polynomial, and all polynomials are continuous on their entire domain. Therefore, \(f(x)\) is continuous on \([0, 1]\), satisfying one of the key conditions of Rolle's Theorem.

Differentiability

A function is differentiable if it has a derivative at every point in an interval. For Rolle's Theorem, the function must be differentiable on the open interval \(a, b\).
Given that \(f(x) = 4x^5 + 3x^3 + 3x - 2\) is a polynomial, it is also differentiable everywhere on its domain. Thus, \(f(x)\) is differentiable on \((0, 1)\).
Earlier, we derived \(f'(x) = 20x^4 + 9x^2 + 3\). Since this derivative is always positive in \(0, 1\), it reinforces the absence of multiple roots within this interval, which links back to the application of Rolle's Theorem.