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

To satisfy the hypotheses of Rolle's Theorem, we need to check:
1. $f(x)$ is continuous on $[0,4]$.
2. $f(x)$ is differentiable on $(0,4)$.
3. $f(0) = f(4)$.
It is easy to verify that $f(x)$ satisfies all three hypotheses of Rolle's Theorem. Now, we just need to find the roots of $f'(x)$ in $(0,4)$ to find the value(s) of $c$ that satisfy the conclusion of Rolle's Theorem.
$$f'(x) = 3x^2 - 10x + 6$$
$$f'(x) = 3(x - \frac{5 + \sqrt{7}}{3})(x - \frac{5 - \sqrt{7}}{3})$$
The roots of $f'(x)$ in $(0,4)$ are $\frac{5 + \sqrt{7}}{3}$ and $\frac{5 - \sqrt{7}}{3}$. Therefore, the value(s) of $c$ that satisfy the conclusion of Rolle's Theorem are $\frac{5 + \sqrt{7}}{3}$ and $\frac{5 - \sqrt{7}}{3}$, which corresponds to choice 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 $f ( x ) = x ^ { 3 } - 5 x ^ { 2 } + 6 x + 1 , [ 0,4 ]$