Verify that the function satisfies the three hypotheses of Rolle's Theorem on the given interval. Then find all numbers c that satisfy the conclusion of Rolle's Theorem.
$$f ( x ) = x ^ { 3 } - 5 x ^ { 2 } + 6 x + 1 , [ 0,4 ]$$
A)
$$c = 2 + \frac { 2 \sqrt { 3 } } { 3 }$$
B)
$$c _ { 1 } = \frac { 5 } { 3 } + \frac { \sqrt { 7 } } { 3 } , c _ { 2 } = \frac { 5 } { 3 } - \frac { \sqrt { 7 } } { 3 }$$
C)
$$c _ { 1 } = 12 + \frac { \sqrt { 2 } } { 3 } , c _ { 2 } = 12 - \frac { \sqrt { 2 } } { 3 }$$
D)
$$c = 2 - \frac { 2 \sqrt { 3 } } { 3 }$$
E)
$$c _ { 1 } = 2 + \frac { 2 \sqrt { 3 } } { 3 } , c _ { 2 } = \frac { 2 \sqrt { 3 } } { 3 }$$

Question

Quizplus · Accepted Answer

Hypothesis 1: f(x) is continuous on [0,4].
The function is a polynomial, so it is continuous everywhere. Therefore, it is continuous on [0,4].

Hypothesis 2: f(x) is differentiable on (0,4).
The derivative of f(x) is f'(x) = 3x^2 - 10x + 6, which is a polynomial and is defined for all x. Therefore, f(x) is differentiable on (0,4).

Hypothesis 3: f(0) = f(4).
f(0) = 1 and f(4) = 1, so f(0) = f(4).

By Rolle's Theorem, there exists at least one number c in (0,4) such that f'(c) = 0. To find c, we need to find the roots of f'(x):
f'(x) = 3x^2 - 10x + 6
= (3x - 2)(x - 3)

Therefore, f'(x) = 0 when x = 2/3 or x = 3. Both of these values are in the interval (0,4), so we have two possible values for c:

c1 = 5/3 + sqrt(7)/3
c2 = 5/3 - sqrt(7)/3

Therefore, the answer is B.

Verify That the Function Satisfies the Three Hypotheses of Rolle's f(x)=x3−5x2+6x+1,[0,4]f ( x ) = x ^ { 3 } - 5 x ^ { 2 } + 6 x + 1 , [ 0,4 ]f(x)=x3−5x2+6x+1,[0,4]

Verify That the Function Satisfies the Three Hypotheses of Rolle's $f ( x ) = x ^ { 3 } - 5 x ^ { 2 } + 6 x + 1 , [ 0,4 ]$