What are the values of c for which Rolle's theorem for the function `f(x)=x^(3)-3x^(2)+2x` in the interval `[0,2]` is verified?

Options

A. `c=+-1`
B. `c=1+-1/(sqrt(3))`
C. `c=+-2`
D. None of these

Answer

B

Explanation

Given `f(x)=x^(3)-3x^(2)+2x`
Here we observe that
a. `f(x)` is a polynomial so it is continuous in the interval `[0,2]`
b. `f'(x)=3x^(2)-6x+2` exists for all ` x epsilon (0,2)`.
so `f(x)` is differentiable for all `x epsilon(0,2)` and
c. `f(0)=0,f(2)=2^(3)-3(2)^(2)+2(2)=0`
`f(0)=f(2)`
Thus, all the three conditions of Rolle's theorem are satisfied.
So, there must exist `c epsilon(0,2)` such that `f'(c)=0`
`implies3c^(2)-6c+2=0`
`=c=6(+-sqrt(36-34))/(2xx3)`
`=1+-1/(sqrt(3)) epsilon(0,2)`