Question

A function \(f(x)\) and interval \([a, b]\) are given. Check if Rolle's Theorem can be applied to fon \([a, b] ;\) if so, find cin \([a, b]\) such that \(f^{\prime}(c)=0\). \(f(x)=x^{2}+x-6\) on [-3,2] .

Short answer

Rolle's Theorem applies, and \(c = -\frac{1}{2}\).

Step-by-step solution

Verify Continuity

For Rolle's Theorem to apply, the function must be continuous on the closed interval \([-3, 2]\). Since \(f(x) = x^2 + x - 6\) is a polynomial, it is continuous everywhere, including the interval \([-3, 2]\).

Verify Differentiability

Next, the function must be differentiable on the open interval \((-3, 2)\). Since derivatives can be found for polynomials everywhere, \(f(x) = x^2 + x - 6\) is differentiable on \((-3, 2)\).

Check Endpoints for Equal Values

Rolle's Theorem requires that \(f(a) = f(b)\). Calculate \(f(-3)\):\[ f(-3) = (-3)^2 + (-3) - 6 = 9 - 3 - 6 = 0 \]Calculate \(f(2)\):\[ f(2) = 2^2 + 2 - 6 = 4 + 2 - 6 = 0 \]Since \(f(-3) = 0\) and \(f(2) = 0\), the selection of endpoints satisfies the condition \(f(a) = f(b)\).

Find \(c\) Where \(f'(c) = 0\)

Since all conditions are met, Rolle's Theorem applies, and there must exist a \(c \) in \((-3, 2)\) such that \(f'(c) = 0\). Compute \(f'(x)\):\[ f'(x) = \frac{d}{dx}(x^2 + x - 6) = 2x + 1 \]Set \(f'(x) = 0\):\[ 2x + 1 = 0 \]Solve for x:\[ 2x = -1 \]\[ x = -\frac{1}{2} \]Thus, \(c = -\frac{1}{2}\) exists in the interval \((-3, 2)\).

Key concepts

Continuity

Continuity is a crucial aspect when applying Rolle's Theorem. A function's continuity on a closed interval means there are no sudden jumps, breaks, or holes. It smoothly connects between points with no interruptions. For the function \(f(x) = x^2 + x - 6\), being a polynomial means it's naturally continuous across all real numbers. Thus, it seamlessly passes through the interval \([-3, 2]\) without any breaks, an essential criterion for Rolle's Theorem to be applicable. In essence:

• No gaps
• No abrupt changes
• Continuous through the interval

Continuity ensures that the function behaves predictably, a requirement when searching for critical points where the derivative equals zero.

Differentiability

Differentiability is another fundamental criterion for using Rolle's Theorem. A function is differentiable on an open interval if it has a derivative at each point within that interval. This means the function has a specific rate of change everywhere inside this range. For our polynomial \(f(x) = x^2 + x - 6\), it's differentiable across any interval, particularly on \((-3, 2)\). The differentiability criterion ensures that we can find derivatives smoothly without any interruptions like sharp corners or cusps. Here,

• Derivatives exist at all points in the open interval
• Smooth changes ensuring polynomial behavior
• No sharp turns or undefined slopes

This smoothness of differentiability is crucial in proving that there exists a point \(c\) where the derivative reaches zero.

Polynomial Functions

Polynomial functions, like \(f(x) = x^2 + x - 6\), are expressions consisting of variables raised to various powers and their coefficients. They exhibit unique simplicity as they are both continuous and differentiable everywhere on their domain of real numbers. This gives them a predictable behavior. There is, plan and steady structure with no unexpected complexity. Some notable traits include:

• Continuous and differentiable over all real numbers
• Smooth curves or lines, no breaks or gaps
• Constructed from simple terms of powers and constants

The nature of polynomial functions makes them ideal for demonstrating the application of Rolle's Theorem. Their characteristics help locate critical points where the slope, or derivative, is zero.

Critical Points

Critical points are where the function reaches a maximum, minimum, or saddle point (neither a max nor min) and are crucial when applying tools like Rolle's Theorem. These points are mathematically identified where the derivative equals zero. For our function \(f(x) = x^2 + x - 6\), establishing a derivative \(f'(x) = 2x + 1\) helps us find these critical points. Set \(f'(x) = 0\) leads us to solve for \(x = -\frac{1}{2}\). This calculation identifies that at \(x = -\frac{1}{2}\), there is a critical point, lying within the interval \((-3, 2)\). Critical points:

• Derivatives become zero
• Intrinsic to the application of Rolle's Theorem
• Help locate peaks or troughs in a function

Finding such points confirms Rolle’s Theorem and the conditions of equal function values at endpoints.