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

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Solution

## The condition for the Mean Value Theorem is given as, (a). The function fis continuous on the close interval [ a,b ]. (b). The function f is differentiable on open interval ( a,b ). (c). The value of function f( a )=f( b ). Then there exists some c∈( a,b ) such that f ′ ( c )= f( b )−f( a ) b−a . (i) The given function is, f( x )=[ x ],x∈[ 5,9 ](1) The first derivative of the function f ′ ( c )=0for 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 function at point 5 is, f( 5 )=[ 5 ] =5 The value of function at point 9 is, f( 9 )=[ 9 ] =9 It can be observed that f( 5 )≠f( 9 ). 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 ]−[ b ] h = lim h→0 b−1−b h = lim h→0 −1 h =∞ The right 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−b h =0 Since, right hand limit is not equal to the left hand limit therefore function is not differentiable in the given interval. Hence, Mean value theorem is not satisfied for the given function f( x )=[ x ]for x∈[ 5,9 ]. (ii) The given function is, f( x )=[ x ],x∈[ −2,2 ](2) From the above equation (1), it is clear that the function f( x )is not continuous at x=−2and 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 )≠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 =∞ The right 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−b h =0 Since, right hand limit is not equal to the left hand limit therefore function is not differentiable in the given interval. Hence, Mean value theorem is not satisfied for the given function f( x )=[ x ] for x∈[ −2,2 ]. (iii) The given function is, f( x )= x 2 −1,x∈[ 1,2 ](3) The first derivative of the function f ′ ( c )=0for some value, where c∈( 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 =1−1 =0 The value of function at point 2 is, f( 2 )= ( 2 ) 2 −1 =4−1 =3 It can be observed that f( 1 )≠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 { b 2 +2bh+ 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. Hence, Mean value theorem is satisfied for the given function f( x )= x 2 −1for x∈[ 1,2 ].

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