the function ( f(x)=\frac{1}{3}x^{3}-25x ) satisfies the hypotheses of rolles theorem on the interval (…

the function ( f(x)=\frac{1}{3}x^{3}-25x ) satisfies the hypotheses of rolles theorem on the interval ( 0,5sqrt{3} ). find all values of ( c ) that satisfy the conclusion of the theorem.\nrolles theorem: let ( f ) be a continuous function on ( a,b ), differentiable on ( (a,b) ), and ( f(a)=f(b) ). then there is a number ( c ) in ( (a,b) ) such that ( f(c)=0 ).\n5\n( pm 5 )\n4\n( pm 4 )\nno correct answer choice is given.

the function ( f(x)=\frac{1}{3}x^{3}-25x ) satisfies the hypotheses of rolles theorem on the interval ( 0,5sqrt{3} ). find all values of ( c ) that satisfy the conclusion of the theorem.\nrolles theorem: let ( f ) be a continuous function on ( a,b ), differentiable on ( (a,b) ), and ( f(a)=f(b) ). then there is a number ( c ) in ( (a,b) ) such that ( f(c)=0 ).\n5\n( pm 5 )\n4\n( pm 4 )\nno correct answer choice is given.

Answer

Explanation:

Step1: Find the derivative of (f(x))

Using the power rule ((x^n)^\prime = nx^{n - 1}), for (f(x)=\frac{1}{3}x^{3}-25x), we have (f^\prime(x)=x^{2}-25).

Step2: Set (f^\prime(c) = 0)

Since (f^\prime(c)=c^{2}-25) and (f^\prime(c) = 0) (by Rolle's Theorem), we solve the equation (c^{2}-25=0). Using the difference - of - squares formula (a^{2}-b^{2}=(a + b)(a - b)), where (a = c) and (b = 5), we get ((c + 5)(c - 5)=0). So (c=-5) or (c = 5). But we are on the interval ((0,5\sqrt{3})\approx(0,8.66)).

Answer:

A. 5