Question

Verify Mean Value Theorem, if in the interval [ a , b ], where a = 1 and b = 3. Find all for which

Answer 1

Consider the function f( x )= x 3 −5 x 2 −3x in the interval [ a,b ], where a=1and b=3.

The above given function is polynomial; therefore, it is a continuous function in [ 1,3 ]and differentiable in the open interval ( 1,3 ).

The first derivative of the function is,

f( x )= x 3 −5 x 2 −3x f ′ ( x )=3 x 2 −10x−3

The value of the function at point 1 is,

f( 1 )= 1 3 −5× 1 2 −3×1 =−7

The value of the function at point 3 is,

f( 3 )= 3 3 −5× 3 2 −3×3 =−27

Further simplify for Mean value theorem.

f( b )−f( a ) b−a = f( 3 )−f( 1 ) 3−1 = { −27−( −7 ) } 2 =−10

According to Mean value theorem, there is a point c∈( 1,3 )such that, f ′ ( c )=−10.

f ′ ( c )=−10 3 c 2 −10c−3=−10 3 c 2 −10c+7=0 3 c 2 −3c−7c+7=0

Further simplify the above equation.

3c( c−1 )−7( c−1 )=0 ( c−1 )( 3c−7 )=0 c=1, 7 3

Where point c= 7 3 ∈( 1,3 )

Hence, Mean Value Theorem is satisfied for the given function.