- Verify that \(g(x)\) is a polynomial, thus continuous and differentiable.
- Evaluate the function at the endpoints to confirm \(g(-2) = g(3) = 0\).
- Calculate the derivative \(g'(x) = 3x^2 + 2x - 8\) and solve \(g'(x) = 0\) to find critical points.
- Identify that only \(c = \frac{4}{3}\) lies within the open interval \((-2, 3)\).
- The final point is \(\boxed{\frac{4}{3}}\).
Explanation
- Verifying Conditions for Rolle's Theorem
To determine if Rolle's Theorem applies to the function \(g(x) = x^3 + x^2 - 8x - 12\) on the interval \([-2, 3]\), we must first verify the three conditions of the theorem:
- The function must be continuous on the closed interval \([-2, 3]\). Since \(g(x)\) is a polynomial, it is continuous for all real numbers, so this condition is met.
- The function must be differentiable on the open interval \((-2, 3)\). As a polynomial, \(g(x)\) is differentiable everywhere, so this condition is met.
- The function values at the endpoints must be equal, i.e., \(g(-2) = g(3)\).
Let's calculate these values:
\(g(-2) = (-2)^3 + (-2)^2 - 8(-2) - 12 = -8 + 4 + 16 - 12 = 0\).
\(g(3) = (3)^3 + (3)^2 - 8(3) - 12 = 27 + 9 - 24 - 12 = 0\).
Since \(g(-2) = g(3) = 0\), all conditions for Rolle's Theorem are satisfied. Therefore, there exists at least one number \(c\) in the open interval \((-2, 3)\) such that \(g'(c) = 0\).
- Finding the Derivative and Critical Points
Having confirmed that Rolle's Theorem applies, we proceed to find the derivative of \(g(x)\):
\(g'(x) = \frac{d}{dx}(x^3 + x^2 - 8x - 12) = 3x^2 + 2x - 8\).
To find the value(s) \(c\) where the slope of the tangent line is zero, we set the derivative equal to zero:
\(3c^2 + 2c - 8 = 0\).
We solve this quadratic equation using the quadratic formula \(c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\(c = \frac{-2 \pm \sqrt{2^2 - 4(3)(-8)}}{2(3)} = \frac{-2 \pm \sqrt{4 + 96}}{6} = \frac{-2 \pm \sqrt{100}}{6} = \frac{-2 \pm 10}{6}\).
This gives two potential values:
\(c_1 = \frac{-2 + 10}{6} = \frac{8}{6} = \frac{4}{3}\)
\(c_2 = \frac{-2 - 10}{6} = \frac{-12}{6} = -2\)
- Identifying Points Guaranteed by Rolle's Theorem
We now examine the results in the context of the open interval \((-2, 3)\).
- The point \(c_1 = \frac{4}{3}\) lies within the interval \((-2, 3)\).
- The point \(c_2 = -2\) does not lie within the open interval \((-2, 3)\), as it is an endpoint.
Thus, the only point guaranteed by Rolle's Theorem is \(c = \frac{4}{3}\). The final answer is \(\boxed{\frac{4}{3}}\).
Examples
Rolle's theorem is a foundational concept in calculus used in physics and engineering to analyze movement. For instance, if an object moves such that its starting position and ending position are the same, Rolle's theorem guarantees there was at least one moment where its instantaneous velocity was zero. This helps engineers design mechanical systems or analyze trajectories where motion must come to a halt or change direction.