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1) it is continuous on the closed interval [a,b],

2) it is differentiable on the open interval (a,b),

3) f(a) = f(b).

Then there exists a real number c $\in$ (a,b) such that f '(c) = 0.

Let f(x) be the real valued function defined on [a,b] such that the curve y =f(x) is a continuous curve between the points A(a, f(a)) and B(b,f(b)) and it is possible to draw a unique tangent at every point on the curve y between A and B. Also, the coordinates of the end points of the interval [a,b] are equal. then, there exists at least one point (c,f(c)) lying between A and B on the curve y = f(x) where tangent is parallel to the X-axis.

Let f(x) be a polynomial function with a and b as its roots. As f(x) is a polynomial function so it is continuous and differentiable. And a and b are the roots of the function so f(a) = f(b) = 0. So f(x) satisfies all the conditions of Rolle's theorem. So there exists c such that c $\in$ (a,b) such that f '(c) = 0 or f'(x) = for x = c.

Between any two roots there is always a root of its derivative f'(x).

1) f(x) = tan(x) on [0,$\pi$]

As we know that the tangent function is not continuous in $\pi$/2 and $\pi$/2 lies in the interval of [0,$\pi$].

So for this function Rolle's theorem is not applicable here.

1) f(x) = $x^{2} - 5x + 6 $ on the interval [2,3]

As the given function is polynomial so it is continuous and differentiable in the interval given.

f(2) = $2^{2} - 5(2) + 6 $

= 4 - 10 + 6

f(2) = 0

f(3) = $3^{2} - 5(3) + 6 $

= 9 - 15 + 6

f(3) = 0

So f(2) = f(3)

As all the conditions of the Rolle's theorem are satisfied so there exists some 'c'.

f(x) = $x^{2} - 5x + 6 $

f'(x) = 2x -5

f'(c) = 2c - 5

0 = 2c - 5

2c = 5

c = 5/2 = 2.5

Hence, Rolle's theorem is verified.

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