Show that f x = 1 x is a decreasing function on 0 , ∞ .

Here, f x = 1 x Let x 1 , x 2 ∈ 0 , ∞ such that x 1 < x 2 . Then, x 1 < x 2 ⇒ 1 x 1 > 1 x 2 ⇒ f x 1 > f x 2 ∴ x 1 < x 2 ⇒ f x 1 > f x 2 , ∀ x 1 , x 2 ∈ 0 , ∞ So, f x is decreasing function on 0 , ∞ .

Show that f x = 1 x is a decreasing function on 0 , ∞ .

Here, f x = 1 x Let x 1 , x 2 ∈ 0 , ∞ such that x 1 < x 2 . Then, x 1 < x 2 ⇒ 1 x 1 > 1 x 2 ⇒ f x 1 > f x 2 ∴ x 1 < x 2 ⇒ f x 1 > f x 2 , ∀ x 1 , x 2 ∈ 0 , ∞ So, f x is decreasing function on 0 , ∞ .

Show that f x = 1 1 + x 2 decreases in the interval [ 0 , ∞ ) and increases in the interval ( − ∞ , 0 ] .

Here, f x = 1 1 + x 2 Case 1 : Let x 1 , x 2 ∈ 0 , ∞ such that x 1 < x 2 . Then, x 1 < x 2 ⇒ x 1 2 < x 2 2 ⇒ 1 + x 1 2 < 1 + x 2 2 ⇒ 1 1 + x 1 2 > 1 1 + x 2 2 ⇒ f x 1 > f x 2 ∀ x 1 , x 2 ∈ 0 , ∞ So, f x is decreasing on 0 , ∞ . Case 2 : Let x 1 , x 2 ∈ ( - ∞ , 0 ] such that x 1 < x 2 Then, x 1 < x 2 ⇒ x 1 2 > x 2 2 ⇒ 1 + x 1 2 > 1 + x 2 2 ⇒ 1 1 + x 1 2 < 1 1 + x 2 2 ⇒ f x 1 < f x 2 ⇒ f x 1 < f x 2 , ∀ x 1 , x 2 ∈ ( - ∞ , 0 ] So, f x is increasing on ( - ∞ , 0 ] .

Show that f x = 1 1 + x 2 is neither increasing nor decreasing on R .

f x = 1 1 + x 2 R can be divided into two intervals 0 , ∞ and ( - ∞ , 0 ] . Case 1 : Let x 1 , x 2 ∈ ( 0 , ∞ ) such that x 1 < x 2 . Then, x 1 < x 2 ⇒ x 1 2 < x 2 2 ⇒ 1 + x 1 2 < 1 + x 2 2 ⇒ 1 1 + x 1 2 > 1 1 + x 2 2 ⇒ f x 1 > f x 2 ∀ x 1 , x 2 ∈ 0 , ∞ So, f x is decreasing on 0 , ∞ . Case 2 : Let x 1 , x 2 ∈ ( - ∞ , 0 ] such that x 1 < x 2 . Then, x 1 < x 2 ⇒ x 1 2 > x 2 2 ⇒ 1 + x 1 2 > 1 + x 2 2 ⇒ 1 1 + x 1 2 < 1 1 + x 2 2 ⇒ f x 1 < f x 2 ∀ x 1 , x 2 ∈ ( - ∞ , 0 ] So, f x is increaing on ( - ∞ , 0 ] . Here, f x is decreasing on 0 , ∞ and increasing on ( - ∞ , 0 ] Thus, f x is neither increasing nor decreasing on R .