Solve Rolle's Theorem for f on [1,3]

[msc8127]

Homework Statement

determine whether rolle's theorem can be applied to f on the closed interval [a,b]. If Rolle's theorem can be applied, find all values of c in the open interval (a,b) such that f'(c) = 0

Homework Equations

f(x) = (x-1)(x-2)(x-3) with the interval being [1,3]

The Attempt at a Solution

the function is continuous on [1,3] and differentiable on (1,3)

f(1) = 0 and f(3) = 0 so f(1) = f(3)

f'(x) = 3x^2 - 12x + 11

now I need to find the point in [1,3] with slope 0 so I set f'(x) = 0.

I know from here I'm supposed to get c values of (6 - sqrt 3) / 3 and (6 + sqrt 3) / 3 however the algebra to get me to this point is eluding me.

Can someone please point of the obvious for me?

Thanks

 

[CompuChip]

It is simply a quadratic equation:
3x^2 - 12x + 11 = 0
You can solve it using your standard toolbox, e.g. by applying the quadrature formula:
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.

 

[brendan]

use use the Quadratic equation to solve for x



x = (-b +/- sqrt(b^2 - 4ac)) / 2a


your equation is f'(x) 3x^2 - 12x + 11

Set for zero

3x^2 - 12x + 11=0


So.

a = 3 b = -12 c= 11

x = (12+/- Sqrt(144 - (4*3*11)))/6

x = (12 +/- Sqrt(12))/6

x = (Sqrt(3)+6)/3 or -(Sqrt(3)-6)/3