The given function is y=log( 1+x )− 2x 2+x ,x>−1.
Differentiate the function with respect to x.
dy dx = 1 ( 1+x ) − ( 2+x )( 2 )−2x ( 2+x ) 2 = 1 ( 1+x ) − 4+2x−2x ( 2+x ) 2 = 1 ( 1+x ) − 4 ( 2+x ) 2 = ( 2+x ) 2 −4x−4 ( 2+x ) 2 ( 1+x )
Further simplify the above expression.
dy dx = 4+ x 2 +4x−4x−4 ( 2+x ) 2 ( 1+x ) = x 2 ( 2+x ) 2 ( 1+x )
Substitute the dy dx = 0.
x 2 ( 2+x ) 2 ( 1+x ) =0 x 2 =0 x=0 ( 2+x ) 2 ≠0,( 1+x )≠0
Now, the point x=0 divides the domain ( −1,∞ ) into two disjoint intervals given by,
−1<x<0 x>0
When, −1<x<0,
x<0⇒ x 2 >0 x>−1⇒( 2+x )>0 ( 2+x ) 2 >0 dy dx >0
When, x>0.
x>0⇒ x 2 >0 ( 2+x ) 2 >0 dy dx >0
Thus, the function y=log( 1+x )− 2x 2+x ,x>−1 is increasing throughout its domain.