SOLUTION: Please help me verify Rolle's theorem for f(x)=(x-1)(x-2)(x-3) in the interval [1,3]

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Question 656696: Please help me verify Rolle's theorem for f(x)=(x-1)(x-2)(x-3)
in the interval [1,3]
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
We need to find a value of x in the interval [1,3]
such that f'(x) = 0.  So we need to find f'(x)

f(x)= (x-1)(x-2)(x-3)

First we'll multiply out the right sides:

f(x)= (x²-3x+2)(x-3)

f(x)= x³-3x²-3x²+9x+2x-6

f(x)= x³-6x²+11x-6

Now we'll take the derivative:

f'(x) = 3x²-12x+11

Now we set the right side equal to zero in 
hopes of getting a value of x in the interval [1,3]

3x²-12x+11 = 0

x = %28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

x = %28-%28-12%29+%2B-+sqrt%28%28-12%29%5E2-4%2A%283%29%2A%2811%29+%29%29%2F%282%2A%283%29%29+

x = %2812+%2B-+sqrt%28144-132%29%29%2F6+

x = %2812+%2B-+sqrt%2812%29%29%2F6+

x = %2812+%2B-+sqrt%284%2A3%29%29%2F6+

x = %2812+%2B-+2sqrt%283%29%29%2F6+

x = %282%286+%2B-+sqrt%283%29%29%29%2F6+

x = %286+%2B-+sqrt%283%29%29%2F3+ 

Using the +,
x = %286+%2B+sqrt%283%29%29%2F3+ ≐ 2.57735 wjich is in the interval [1,3] 

Using the -,
x = %286+-+sqrt%283%29%29%2F3+ ≐ 1.42265 whis is also in the interval [1,3]

So there are two values of x in the interval [1,3] where f'(x) = 0.

[There only needs to be one such value to satisfy Rolle's theorem,
but it is just fine if there are more than one.]

Edwin