Examine if Rolle’s Theorem is applicable to any of the following functions. Can you say some thing about the converse of Rolle’s Theorem from these examples? (i) (ii) (iii)
(i)
The given function is,
f ( x ) = [ x ] , x ∈ [ 5 , 9 ] (1)
The condition for the Rolle’s Theorem given is,
(a) The function f is continuous on the close interval [ a , b ] .
(b) The function f is differentiable on open interval ( a , b ) .
(c) f ( a ) = f ( b ) .
The first derivative of the function f ′ ( c ) = 0 for some value, where c ∈ ( a , b ) .
From the above equation (1), it is clear that the function f ( x ) is not continuous at x = 5 and x = 9 .
The value of the function is calculated as follows,
f ( 5 ) = [ 5 ] = 5
And,
f ( 9 ) = [ 9 ] = 9
Therefore, f ( 5 ) ≠ f ( 9 ) .
The differentiability of the function is checked as follows.
Consider b is an integer.
The left hand limit of f at x = b is,
lim h → 0 f ( b + h ) − f ( b ) h = lim h → 0 [ b + h ] − [ b ] h = lim h → 0 b − 1 − b h = lim h → 0 − 1 h = ∞
The right hand limit of f at x = b is,
lim h → 0 f ( b + h ) − f ( b ) h = lim h → 0 [ b + h ] − [ b ] h = lim h → 0 b − b h = 0
The right hand limit is not equal to the left hand limit; therefore, function f is not differentiable in the given interval.
Hence, Rolle’s Theorem is not satisfied for the given function.
(ii)
The given function is,
f ( x ) = [ x ] , x ∈ [ − 2 , 2 ] (1)
From the above equation (1), it is clear that the function f ( x ) is not continuous at x = − 2 and x = − 2 .
The value of the function is calculated as follows.
f ( − 2 ) = [ − 2 ] = − 2
And,
f ( 2 ) = [ 2 ] = 2
The above equation shows that value of f ( − 2 ) ≠ f ( 2 ) .
The differentiability of the function is calculated as follows.
Consider b is an integer.
The left hand limit of f at x = b is,
lim h → 0 f ( b + h ) − f ( b ) h = lim h → 0 [ b + h ] − [ b ] h = lim h → 0 b − 1 − b h = lim h → 0 − 1 h = ∞
The right hand limit of f at x = b is,
lim h → 0 f ( b + h ) − f ( b ) h = lim h → 0 [ b + h ] − [ b ] h = lim h → 0 b − b h = 0
The right hand limit is not equal to the left hand limit; therefore, function f is not differentiable in the given interval.
Hence, Rolle’s Theorem is not satisfied for the given function.
(iii)
The given function is,
f ( x ) = x 2 − 1 , x ∈ [ 1 , 2 ] (3)
The above polynomial equation shows that the given function continuous and differentiable in the given interval.
The value of the function is calculated as follows,
f ( x ) = ( x ) 2 − 1 f ( 1 ) = ( 1 ) 2 − 1 = 0
Further value calculated at f ( 2 ) ,
f ( 2 ) = ( 2 ) 2 − 1 = 4 − 1 = 3
Therefore, f ( 1 ) ≠ f ( 2 )
Therefore, the above calculation shows that the Rolle’s Theorem is not satisfied for the given function.
(i)
The given function is,
The condition for the Rolle’s Theorem given is,
(a) The function
(b) The function
(c)
The first derivative of the function
From the above equation (1), it is clear that the function
The value of the function is calculated as follows,
And,
Therefore,
The differentiability of the function is checked as follows.
Consider
The left hand limit of
The right hand limit of
The right hand limit is not equal to the left hand limit; therefore, function f is not differentiable in the given interval.
Hence, Rolle’s Theorem is not satisfied for the given function.
(ii)
The given function is,
From the above equation (1), it is clear that the function
The value of the function is calculated as follows.
And,
The above equation shows that value of
The differentiability of the function is calculated as follows.
Consider
The left hand limit of
The right hand limit of
The right hand limit is not equal to the left hand limit; therefore, function
Hence, Rolle’s Theorem is not satisfied for the given function.
(iii)
The given function is,
The above polynomial equation shows that the given function continuous and differentiable in the given interval.
The value of the function is calculated as follows,
Further value calculated at
Therefore,
Therefore, the above calculation shows that the Rolle’s Theorem is not satisfied for the given function.
