Examine the applicability of Mean Value Theorem for all three functions given in the above exercise 2.

Explanation

(i) As the function `f (x) = [x]` is neither continuous nor derivable , conditions of mean value theorem is not satisfied hence not verified .
(ii) Given function is,
`f( x )=[x], x in [−2,2 ]`
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 function at point `−2` is,
`f( −2 )=[ −2 ] =−2`
The value of function at point `2` is,
`f( 2 )=[ 2 ] =2`
It can be observed that `f( −2 )`in`f( 2 )`.
The differentiability of the function can be check as follows. Let, b is an integer.
The left hand limit of fat `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 =oo`
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`
Since, right hand limit is not equal to the left hand limit therefore function is not differentiable in the given interval.
Therefore, Mean value theorem is not satisfied for the given function `f( x )=[x]` for `x in [−2,2]`.
(iii)Given function is,
`f(x)= x^2 −1 , x in [1,2]`
The first derivative of the function `f′(c)=0` for some value, where `c in (a,b).`
From the above equation (1), it is clear that the function `f(x)` is continuous at `x=1 and x=2`.
The value of function at point 1 is,
`f(1)= (1)^2−1=0`
The value of function at point 2 is,
`f(2)= (2)^2−1=3`
It can be observed that `f(1) in f(2)`.
The differentiability of the function can be check as follows. Let, b is an integer.
The left hand limit of the function at `x=b` is,
`lim_(h->0)(f(b+h)−f b))/h = lim_(h->0) ({(b+h)^2−1}−{ b^2−1})/h = lim_(h->0)({ b^2 +2bh+ h^2−1}− (b^2 +1))/h = lim_(h->0) (2bh)/ h =2b`
The right hand limit of fat x=b is,
`lim_(h->0)(f( b+h )−f( b ))/h = lim_(h->0)({( b+h)^2−1 }−{ b^2 −1})/ h = lim_(h->0) 2bh/h =2b`
Since, right hand limit is equal to the left hand limit therefore function is differentiable in the given interval.
Therefore, Mean value theorem is satisfied for the given function `f( x )= x^2−1` for `x in[ 1,2 ]`.