If a function is:
Continuous on [a, b].
Differentiable on (a, b).
f(a) = f(b).
Then, there exists a point, c (a, b) such that f'(c) = 0.
The graphical interpretation of Rolle's Theorem states that there is a point where the tangent is parallel to the x-axis.
1. Given the function , determine if Rolle's Theorem is varified on the interval [0, 3]?
First, verify that the function is continuous at x = 1.
Secondly, check if the function is differentiable at x = 1.
The function is not differentiable on the interval (0, 3) and therefore does not satisfy Rolle's Theorem.
2.Is Rolle's Theorem applicable to the function f(x) = ln (5 − x2) on the interval [−2, 2]?
First, calculate the domain of the function.
The function is continuous on the interval [−2, 2] and differentiable on (−2, 2), because the intervals are contained in .
Also, it is determined that f(−2) = f(2), therefore Rolle's theorem is applicable to this function.
3.Verify that the equation x7 + 3x + 3 = 0 has only one real solution.
The function f(x) = x7 + 3x + 3 is continuous and differentiable at ·
f(−1) = −1
f(0) = 3
The equation has at least one solution in the interval (-1, 0).
f' (x) = 7x6 + 3
As the derivative is not annulled in any value, it contradicts Rolle's theorem and therefore only has one real root.