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Question

Verify the Rolle's theorem for the function f(x)=(x1)(x2)(x3),x[1,4]

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Solution

Now, Rolle's theorem states that, for a differentiable function, between two consecutive same valued points, there must be at least one stationary point in between i.e.
The function f(x)=(x1)(x2)(x3) x[1,4] has same value at x=1, x=2, x=3 i.e. it is zero.
Hence, it can be divided into 2 intervals x[1,2] and x[2,3].
f(c)=0 where c[a,b] f(a)=f(b).
f(x)=(x1)(x2)+(x1)(x3)+(x2)(x3)
Now, put f(x)=0.
(x1)(x2)+(x1)(x3)+(x2)(x3)=0
3x212x+11=0
x=12±1224×3×112×3
x=12±126
x=2±13
x=2+13[2,3] and x=213[1,2]
Hence proved.

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