The function $f ( x ) = \frac { 1 } { 3 } x ^ { 3 } - 25 x$ satisfies the hypotheses of Rolle's Theorem on the interval $[ 0,5 \sqrt { 3 } ]$. Find all values of $c$ that satisfy the conclusion of the theorem.
A) $\pm 5$
B) 5
C) $\pm 4$
D) 4

Question

Quizplus · Accepted Answer

The derivative of $f(x)$ is $f'(x) = x^2 - 25$. Since $f'(x)$ is continuous on $[0, 5\sqrt{3}]$ and $f'(0) = f(5\sqrt{3}) = 0$, by Rolle's Theorem, there exists a number $c$ in $(0, 5\sqrt{3})$ such that $f'(c) = 0$. Solving $f'(c) = 0$ gives $c = \pm 5$.

The Function f(x)=13x3−25xf ( x ) = \frac { 1 } { 3 } x ^ { 3 } - 25 xf(x)=31​x3−25x

The Function $f ( x ) = \frac { 1 } { 3 } x ^ { 3 } - 25 x$