SOLVING PROBLEMS USING ROLLE'S THEOREM

Let f (x) be continuous on a closed interval [a, b] and differentiable on the open interval (a, b)

If f(a)  =  f(b)

then there is at least one point c ∈ (a,b) where f '(c) = 0.

Geometrically this means that if the tangent is moving along the curve starting at x = a towards x = b then there exists a c ∈ (a, b) at which the tangent is parallel to the x -axis.

Using the Rolle’s theorem, determine the values of x at which the tangent is parallel to the x -axis for the following functions :

(i) f(x)  =  x² − x, x ∈ [0, 1]

Solution :

f(x) is defined and continuous on the closed interval [0, 1] and it is differentiable on (0, 1).

(1)  =  (2)

So, there is at least one point c ∈ [0, 1]

f'(c) =  2x-1

2c-1  =  0

2c  =  1

c  =  1/2

(ii)  f(x)  =  (x²-2x)/(x+2), x ∈ [-1, 6]

Solution :

f(x) is defined and continuous on the closed interval [-1, 6] and it is differentiable on (-1, 6).

(1)  =  (2)

So, there is at least one point c ∈ [-1, 6]

f'(x)  =  [(x+2) (2x-2) - (x²-2x) (1)] / (x+2)²

f'(x)  =  (2x²+2x-4 - x²+2x) / (x+2)²

f'(x)  =  (x²+4x-4) / (x+2)²

f'(x)  =  0

x²+4x-4  =  0

So, the solutions are -2±2√2.

(iii)  f(x)  =   √x - (x/3), x∈ [0, 9]

Solution :

f(x) is defined and continuous on the closed interval [0, 9] and it is differentiable on (0, 9).

(1)  =  (2)

So, there is at least one point c ∈ [-1, 6]

f(x)  =   √x - (x/3)

f'(x)  =   1/2√x - (1/3)

1/2√x - (1/3)  =  0

1/2√x  =  1/3

3  =  2√x

√x  =  3/2

x  =  9/4 ∈ [-1, 6]

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