Question

Verify Rolle’s Theorem for the function

Answer 1

Let, f( x )= x 2 +2x−8.

The given polynomial function is continuous in close interval [ −4,2 ]and differentiable in open interval ( −4,2 ).

The value of function at point −4 is,

f( x )= x 2 +2x−8 f( −4 )={ ( −4 ) 2 +2×( −4 )−8 } ={ 16−8−8 } =0

The value of function at point 2 is,

f( 2 )={ ( 2 2 )+2×2−8 } ={ 4+4−8 } =0

The value of the function f( −4 )=f( 2 ) coincides.

According to Rolle’s Theorem, there is a point c∈( −4,2 ) in such a way that the first derivative of the function is zero.

f( x )= x 2 +2x−8 f ′ ( x )=2x+2 f ′ ( c )=o 2c+2=0

Further simplify the above equation,

2c+2=0 c=−1

Where c is equal to −1∈( −4,2 ).

Hence, for the given function Rolle’s Theorem is verified.