Problem 25 - Calculus Single Variable

Question

In each of Exercises \(23-28\), verify that the hypotheses of the Rolle's Theorem hold for the given function \(f\) and interval I. The theorem asserts that \(f^{\prime}(c)=0\) for some \(c\) in \(I .\) Find such a \(c\). $$ x-x^{3 / 2}, \quad I=[0,1] $$

Step-by-Step Solution

Answer

Rolle's Theorem applies; \(c = \frac{4}{9}\).

Step 1: Check Continuity

Rolle's Theorem requires the function to be continuous on the closed interval \([a, b]\). Here, the function is \(f(x) = x - x^{3/2}\). This function is continuous for all \(x \geq 0\), as both \(x\) and \(x^{3/2}\) are continuous functions on \([0, 1]\). So, \(f(x)\) is continuous on \([0, 1]\).

Step 2: Check Differentiability

The function must be differentiable on the open interval \((a, b)\). For \(f(x) = x - x^{3/2}\), the derivative is \(f'(x) = 1 - \frac{3}{2}x^{1/2}\). This derivative exists for all \(x \geq 0\), hence \(f(x)\) is differentiable on \((0, 1)\).

Step 3: Check Equal Function Values at Endpoints

For Rolle's Theorem, \(f(a)\) must equal \(f(b)\) where \([a,b]=[0,1]\). We calculate: \(f(0) = 0^1 - 0^{3/2} = 0\) and \(f(1) = 1 - 1^{3/2} = 0\). Since \(f(0) = f(1) = 0\), this condition is satisfied.

Step 4: Apply Rolle's Theorem

By Rolle's Theorem, there must be some \(c\) in \((0, 1)\) where \(f'(c) = 0\). We solve for \(c\): \(f'(x) = 1 - \frac{3}{2}x^{1/2} = 0\). Thus, \(1 = \frac{3}{2} \sqrt{x}\), leading to \(\sqrt{x} = \frac{2}{3}\), i.e., \(x = \left(\frac{2}{3}\right)^2 = \frac{4}{9}\).

Step 5: Verify \(c\) Lies in the Interval

We found \(c = \frac{4}{9}\). Check if \( \frac{4}{9} \in (0, 1)\). Since \(\frac{4}{9} \approx 0.44\), it clearly lies in \((0, 1)\). Thus, all conditions of Rolle's Theorem are satisfied and \(c = \frac{4}{9}\) is valid.

Key Concepts

Continuity

In calculus, the concept of continuity is crucial to understanding how functions behave across intervals. Continuity means that a function doesn't have any sudden jumps or breaks in its values. For Rolle's Theorem, the function must be continuous on the closed interval \[a, b\].

When addressing the exercise with the function \(f(x) = x - x^{3/2}\), we state that this function is continuous on \([0, 1]\). This is because both parts of the function, \(x\) and \(x^{3/2}\), are continuous on this interval. Continuous functions can be easily identified as those which you can draw without lifting your pencil. This seamless connectivity ensures the first requirement of Rolles’ Theorem is met.

Differentiability

Differentiability ensures that a function has a defined tangent at each point in the interior of the interval. For the function to be differentiable on an interval \(a, b\), the derivative must exist. This is a bit stricter than continuity because a function can be continuous but not differentiable.

In the problem given, the derivative of \(f(x) = x - x^{3/2}\) is calculated as \(f'(x) = 1 - \frac{3}{2}x^{1/2}\). As long as we can compute this derivative across the open interval \((0, 1)\), the function satisfies the differentiability criterion for Rolle’s Theorem. Showing that the derivative is valid on this interval confirms differentiability.

Function Values at Endpoints

A key condition of Rolle's Theorem is that the function values at the endpoints of the interval must be equal, i.e., \(f(a) = f(b)\).

For our problem, we need to compare the function values at 0 and 1:

• Calculate \(f(0) = 0^1 - 0^{3/2} = 0\).
• Calculate \(f(1) = 1 - 1^{3/2} = 0\).

Since both \(f(0)\) and \(f(1)\) equal 0, this condition is satisfied. This shows that the graph of the function does not rise or fall as it moves from the start to the end of the interval.

Derivative

The derivative represents the rate of change or the slope of the tangent line to the curve of a function. It plays a vital role in Rolle's Theorem as it helps identify points on the function where the slope is zero.

For the function \(f(x) = x - x^{3/2}\), its derivative \(f'(x) = 1 - \frac{3}{2}x^{1/2}\) is zero when the function’s tangent is horizontal. Solving \(1 - \frac{3}{2}\sqrt{x} = 0\), we find \(\sqrt{x} = \frac{2}{3}\), meaning \(x = \left(\frac{2}{3}\right)^2 = \frac{4}{9}\). The expression \(f'(x) = 0\) confirms where the rate of change is zero, indicating a potential maximum, minimum, or a point of inflection.

Intervals in Calculus

Intervals define the sections of the x-axis over which we consider the behavior of functions. Understanding intervals is essential in calculus to apply theorems like Rolle's Theorem properly.

In this exercise, our interval is \([0, 1]\), a closed interval where we examine the endpoints, and \((0, 1)\), an open interval used when discussing differentiability. The intervals signify where the function's specific properties, like continuity and differentiability, need checking. Given \(c = \frac{4}{9}\) falls in \((0, 1)\), all conditions of Rolle's Theorem on the interval are verified. Correctly identifying and working with these intervals ensures the accurate application of calculus principles.