Problem 50 - Calculus Single Variable

Question

In Exercises \(49-52\) find a value \(c\) the existence of which is guaranteed by Rolle's Theorem applied to the given function \(f\) on the given interval \(I\). $$ f(x)=x^{3}-x, \quad I=[0,1] $$

Step-by-Step Solution

Answer

The value of \(c\) is \(\sqrt{\frac{1}{3}}\) on the interval \([0, 1]\).

Step 1: Verify Conditions for Rolle's Theorem

Rolle's Theorem requires that the function \(f\) be continuous on the closed interval \(I\) and differentiable on the open interval \((a, b)\). Additionally, \(f(a) = f(b)\). For the given function \(f(x) = x^3 - x\) on \(I = [0, 1]\): \(f(x)\) is a polynomial, thus continuous and differentiable on all real numbers, satisfying these conditions for the interval. Now, evaluate \(f(0)\) and \(f(1)\): - \(f(0) = 0^3 - 0 = 0\) - \(f(1) = 1^3 - 1 = 0\) Thus, \(f(a) = f(b) = 0\) and all conditions for Rolle's Theorem are satisfied.

Step 2: Find the Derivative of the Function

To find the value of \(c\), first determine the derivative \(f'(x)\). Differentiate \(f(x) = x^3 - x\): \[ f'(x) = \frac{d}{dx}(x^3) - \frac{d}{dx}(x) = 3x^2 - 1 \] This derivative function will be used to find the value of \(c\), where \(f'(c) = 0\).

Step 3: Solve for \(c\) Using the Derivative

Set the derivative \(f'(x)\) equal to zero and solve for \(x\): \[ 3x^2 - 1 = 0 \] \[ 3x^2 = 1 \] \[ x^2 = \frac{1}{3} \] \[ x = \pm \sqrt{\frac{1}{3}} \] Since \(x\) must be in the interval \([0, 1]\), the negative solution is not applicable. Therefore, \(c = \sqrt{\frac{1}{3}}\).

Key Concepts

Derivatives

Understanding derivatives is essential when working with topics like Rolle's Theorem. A derivative represents the rate at which a function is changing at any given point. In simpler terms, it gives you the slope of the tangent line to the function's graph at a specific point. To find the derivative of a function, you differentiate it. This involves applying rules like the power rule, product rule, and chain rule, depending on the type of function you have. For the function \( f(x) = x^3 - x \), the derivative \( f'(x) \) is found by applying the power rule. This rule involves bringing down the power as a coefficient and reducing the power by one:

• From \( x^3 \), you get \( 3x^2 \)
• From \( x \), you get \( 1 \) (as the derivative of \( x \) is 1)

Thus, the derivative \( f'(x) = 3x^2 - 1 \). We use this to find points where the slope is zero, which is crucial for applying Rolle's Theorem.

Polynomial Functions

Polynomial functions, like \( f(x) = x^3 - x \), are expressions involving variables raised to powers and combined using addition or subtraction coefficients. They are an important class of functions in calculus and mathematics in general.Polynomials can range from simple (linear functions) to more complex forms (such as cubic, quartic, etc.). The degree of a polynomial is determined by the highest power of the variable present in the function. In our case, \( f(x) = x^3 - x \) is a cubic polynomial since the highest power of \( x \) is 3.Polynomials are well-behaved functions because their graphs are smooth and continuous. This property comes in handy when applying theorems in calculus that require continuity, such as Rolle's Theorem. Additionally, polynomial functions are differentiable everywhere, which further aids in their analysis in calculus.

Continuous Functions

Continuity is a fundamental concept in calculus, particularly when applying theorems like Rolle's Theorem. A function is continuous over an interval if there are no breaks, jumps, or holes in its graph within that interval.Continuous functions allow for smooth transitions from one point to another without interruptions. In mathematical terms, if a function \( f(x) \) is continuous at a point \( x = a \), the limit of \( f(x) \) as \( x \) approaches \( a \) is equal to \( f(a) \).With Rolle's Theorem, continuity ensures that, along a continuous path from \( x = 0 \) to \( x = 1 \), the function \( f(x) \) behaves predictably. If \( f(x) \) is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), with the condition \( f(a) = f(b) \), then there must be at least one point \( c \) within \( (a, b) \) where the derivative \( f'(c) = 0 \). This is how we found the value of \( c = \sqrt{\frac{1}{3}} \) for our function.