Problem 15 - Calculus of a Single Variable

Question

Determine whether Rolle's Theorem can be applied to \(f\) on the closed interval \([a, b] .\) If Rolle's Theorem can be applied, find all values of \(c\) in the open interval \((a, b)\) such that \(f^{\prime}(c)=0 .\) If Rolle's Theorem cannot be applied, explain why not. \(f(x)=\frac{x^{2}-2 x-3}{x+2}, \quad[-1,3]\)

Step-by-Step Solution

Answer

Rolle's Theorem cannot be applied to \( f(x) = \frac{x^{2}-2 x-3}{x+2} \) on the given interval \([-1,3]\) because the function does not have the same value at both ends of this interval.

Step 1: Check the continuity and differentiability of the function

Firstly, we need to identify any points of discontinuity. \( f(x) \) is undefined when the denominator is zero, i.e., \( x = -2 \). However, \( -2 \) is not in the interval \([-1,3]\), so we can ignore it. The function \( f(x) = \frac{x^{2}-2 x-3}{x+2} = x - 4 + \frac{8}{x + 2} \) is continuous and differentiable on the interval \([-1, 3]\) except at \( x = -2 \). Hence, we can apply Rolle's Theorem to it on the given closed interval.

Step 2: Check if f(a) equals f(b)

Rolle's Theorem only applies if the function has the same y-value at the end points of the interval. Our function \( f(x) \) translates to \( f(x) = x - 4 + \frac{8}{x + 2} \). This gives us \( f(-1) = -1 - 4 + \frac{8}{-1 + 2} = -5 + 8 = 3 \) and \( f(3) = 3 - 4 + \frac{8}{3 + 2} = -1 + \frac{8}{5} \), which do not equal to each other. So, we cannot apply Rolle's Theorem.

Step 3: Conclusion

Given the conditions of Rolle's Theorem, there is no value of \( c \) in the open interval \( (-1,3) \) such that \( f'(c) = 0 \), because \( f(a) \) is not equal to \( f(b) \).

Key Concepts

Continuity

Continuity is a fundamental concept in calculus, especially when applying Rolle’s Theorem. It means that a function doesn’t have any breaks, jumps, or holes in its graph. For Rolle’s Theorem to be applicable, the function must be continuous over the entire closed interval \([a, b]\).

In our exercise, the function \( f(x) = \frac{x^2 - 2x - 3}{x+2} \) needs to be continuous on \([-1, 3]\). A key point for continuity is ensuring there are no points of discontinuity, such as where the denominator becomes zero. Here, \( x = -2 \) makes the denominator zero, but since \( -2 \) is not within our interval, \( f(x) \) remains continuous on \([-1, 3]\). Hence, the first condition of Rolle's Theorem is satisfied.

Differentiability

Differentiability ensures that the function has a well-defined derivative at every point in the open interval \((a, b)\). Simply put, the graph of the function must be smooth without any sharp corners or cusps.

For \( f(x) = \frac{x^2 - 2x - 3}{x + 2} \), we need to confirm that it can be differentiated smoothly in \((-1, 3)\). Since there are no sharp turns or vertical tangents within this region, \( f(x) \) is indeed differentiable.

This condition aligns perfectly with Rolle's Theorem requirements, along with the continuity condition we've already verified.

Closed Interval

A closed interval, noted as \([a, b]\), includes both endpoints \(a\) and \(b\). This is important for Rolle’s Theorem, as we must evaluate the function at the endpoints to test its applicability.

In our case, the closed interval is \([-1, 3]\), meaning \( f(x) \) must be examined at both \( x = -1 \) and \( x = 3 \). While checking for continuity and differentiability, it's crucial to ensure these properties apply over the whole closed interval, not just its interior. Fortunately, we've clarified that our function doesn’t have any discontinuities or non-differentiable points within this closed interval.

Endpoints Equality

Rolle's Theorem has one more essential condition: the function must have equal values at the endpoints of the interval. This means \( f(a) = f(b) \).

In this exercise, we compute \( f(-1) = 3 \) and \( f(3) = -1 + \frac{8}{5} \). Clearly, these values are not equal, which directly tells us that Rolle's Theorem cannot be applied in its full capacity here.

This endpoint inequality shows why we can’t find any \( c \) in the interval \((-1, 3)\) where \( f'(c) = 0 \). Without equal values at the endpoints, the symmetry required for Rolle's Theorem doesn’t exist.