Question

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

Answer 1

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!

Answer 2

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]}