Consider the function f(x) = x^3-3x^2+2x-18 on the interval [ 0 , 2 ]. Verify that this function satisfies the three hypotheses of Rolle's Theorem on the inverval:

f(x) is ___on[0,2]
f(x) is ___ on (0,2)
and f(0)=f(2)=____
then by rolle's theorem, there exists a c such that f'(c)=0.
c1=____ and c2=_____withc1<c2

It is continuous

derivative = 3x^2-6x+2 which is fine
f(0)=-18
f(2) = 8-12+4-18 = -18
find zeros of 3x^2-6x+2 = 0
(3-sqrt3)/3 and (3+sqrt3 )/3

To verify that the function satisfies the hypotheses of Rolle's Theorem on the interval [0, 2], we need to check the following conditions:

1. f(x) is continuous on [0, 2]: A function is continuous on an interval if it is continuous at every point within that interval. In this case, the function f(x) = x^3 - 3x^2 + 2x - 18 is a polynomial function, and polynomial functions are continuous over their entire domain, including the interval [0, 2]. Therefore, f(x) is continuous on [0, 2].

2. f(x) is differentiable on (0, 2): A function is differentiable on an open interval (a, b) if its derivative exists at every point within that interval. To check this condition, we need to find the derivative of f(x). Taking the derivative of f(x) = x^3 - 3x^2 + 2x - 18, we get f'(x) = 3x^2 - 6x + 2. The derivative exists for all real numbers, so f(x) is differentiable on (0, 2).

3. f(0) = f(2): We need to evaluate the function at the endpoints of the interval [0, 2]. Plugging in x = 0, we get f(0) = (0)^3 - 3(0)^2 + 2(0) - 18 = -18. Similarly, evaluating at x = 2, we get f(2) = (2)^3 - 3(2)^2 + 2(2) - 18 = -10. Therefore, f(0) = -18 and f(2) = -10.

Now, since the function satisfies all the hypotheses of Rolle's Theorem, we can conclude that there exists at least one value c in the interval (0, 2) such that f'(c) = 0. To find the values of c, we need to solve the equation f'(c) = 0.

The derivative of f(x) is f'(x) = 3x^2 - 6x + 2. Setting it to zero and solving for x:

3x^2 - 6x + 2 = 0

Using the quadratic formula, x = (-b ± sqrt(b^2 - 4ac)) / (2a),
where a = 3, b = -6, and c = 2:

x = (-(-6) ± sqrt((-6)^2 - 4*3*2)) / (2*3)
= (6 ± sqrt(36 - 24)) / 6
= (6 ± sqrt(12)) / 6
= (6 ± 2sqrt(3)) / 6

Simplifying further, we have:
c1 = (6 + 2sqrt(3)) / 6 = (1 + sqrt(3)) / 3
c2 = (6 - 2sqrt(3)) / 6 = (1 - sqrt(3)) / 3

Therefore, c1 = (1 + sqrt(3)) / 3 and c2 = (1 - sqrt(3)) / 3 are the values of c that satisfy the conclusion of Rolle's Theorem. Additionally, c1 < c2 since (1 + sqrt(3)) / 3 > (1 - sqrt(3)) / 3.