Question

Let I be any interval disjoint from (−1, 1). Prove that the function f given by is strictly increasing on I .

Answer 1

The given function f is defined as,

f( x )=x+ 1 x

The derivative of function f is given as,

f ′ ( x )= df( x ) dx = d( x+ 1 x ) dx =1− 1 x 2

Determine the disjoint in the given interval by,

f ′ ( x )=0 1− 1 x 2 =0 x=±1

From the above equation, x=1 and x=−1. Divide the real axis in three disjoint intervals, ( −∞,1 ),( −1,1 ) and ( 1,∞ ).

For the interval ( −1,1 ),

−1<x<1 x 2 <1 1< 1 x 2 , x≠0 1− 1 x 2 <0, x≠0

Therefore, f ′ ( x )<0 in interval ( −1,1 ). Hence, function f is strictly decreasing in ( −1,1 ).

For the intervals ( −∞,−1 ) and ( 1,∞ ),

x<−1 1<x 1> 1 x 2 1− 1 x 2 >0

Therefore, f ′ ( x )>0 in intervals ( −∞,1 ) and ( 1,∞ ). Hence, function f is strictly increasing in intervals ( −∞,1 ) and ( 1,∞ ).

Therefore, it is proved that the function f is strictly increasing in the given interval I disjoint from ( −1,1 ).