Show that f ( x ) = sin x − cos x is an increasing function on ( − π 4 , π 4 ) ?

Given that f x = sin x - cos x f ' x = cos x + sin x = cos x 1 + sin x cos x = cos x 1 + cot x Here, - π 4 < x < π 4 ⇒ cos x > 0 . . . 1 Also, - π 4 < x < π 4 ⇒ - 1 < cot x < 1 ⇒ 0 < 1 + cot x < 2 ⇒ 1 + cot x > 0 . . . 2 cos x 1 + cot x > 0 , ∀ x ∈ - π 4 , π 4 From eqs . ( 1 ) and ( 2 ) ⇒ f ' x > 0 , ∀ x ∈ - π 4 , π 4 So, f ( x ) is increasing interval - π 4 , π 4 .

Determine whether f ( x ) = − x 2 + sin x is increasing or decreasing on ( − π 3 , π 3 ) ?

Given that f x = - x 2 + sin x f ' x = - 1 2 + cos x Here, - π 3 < x < π 3 ⇒ cos x > 1 2 ⇒ - 1 2 + cos x > 0 ⇒ f ' x > 0 , ∀ x ∈ - π 3 , π 3 So, f ( x ) is increasing on - π 3 , π 3

Find the intervals in which f ( x ) = log ( 1 + x ) - x 1 + x is increasing or decreasing ?

f x = log 1 + x - x 1 + x Domain of f ( x ) is ( - 1 , ∞ ) f ' x = 1 1 + x - 1 + x - x 1 + x 2 = 1 1 + x - 1 1 + x 2 = x 1 + x 2 For f ( x ) to be increasing, we must have f ' x > 0 ⇒ x 1 + x 2 > 0 ⇒ x > 0 ∵ 1 + x 2 > 0 , Domain : - 1 , ∞ ⇒ x ∈ 0 , ∞ So, f ( x ) is increasing on ( 0 , ∞ ) For f ( x ) to be decreasing, we must have f ' x < 0 ⇒ x 1 + x 2 < 0 ⇒ x < 0 ∵ 1 + x 2 > 0 , Domain : - 1 , ∞ ⇒ x ∈ - 1 , 0 So, f ( x ) is decreasing in the interval ( - 1 , 0 )