Verify Rolle theorem for function: f(x)= #x^3-6x^2+11x-6# in the interval [1,3] ?

2 Answers
Apr 5, 2018

Since #f(1) = f(3) =0#, and #f(x)# is continuous on #[1, 3]#, there must be a value of #c# on #[1, 3]# where #f'(c) = 0#.

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

#0 = 3c^2 - 12c + 11#

#c = (12 +- sqrt((-12)^2 - 4 * 3 * 11))/(2 * 3)#

#c = (12 +- sqrt(12))/6#

#c = (12 +- 2sqrt(3))/6#

#c = 2 +- 1/3sqrt(3)#

Using a calculator we get

#c ~~ 1.423 or 2.577#

Since these are within #[1, 3]# this confirms Rolle's Theorem.

Hopefully this helps!

Apr 5, 2018

See the explanation below.

Explanation:

Rolle's theorem states that if a function #f(x)# is continuous on the interval #[a,b]# and differentiable on the interval #(a,b)# and if #f(a)=f(b)# then there exists #c in (a,b)# such that

#f'(c)=0#

Here,

#f(x)=x^3-6x^2+11x-6#

The interval is #I=(1,3)#

#f(1)=1^3-6xx1^2+11xx1-6=0#

#f(3)=3^3-6xx3^2+11xx3-6=0#

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

Therefore,

#f'(c)=3c^2-12c+11=0#

Solving this quadratic equation in #c#

#c=(12+-sqrt(144-4*3*11))/6#

#=(12+-sqrt12)/6#

#c_1=2.85#

#c_2=1.42#

#c_1# and #c_2# are in the interval #(1,3)#

Therefore,
Rolle theorem is verified.

graph{x^3-6x^2+11x-6 [-2.335, 5.46, -1.673, 2.222]}