Find the values of x for which f x = x x 2 + 1 is (a) strictly increasing (b) strictly decreasing

The given function is, f x = x x 2 + 1 . Differentiating w.r.t. x , we get f ' x = d d x x x 2 + 1 f ' x = x 2 + 1 . d d x x - x d d x x 2 + 1 x 2 + 1 2 f ' x = x 2 + 1 1 - x 2 x + 0 x 2 + 1 2 f ' x = x 2 + 1 - 2 x 2 x 2 + 1 2 f ' x = 1 - x 2 x 2 + 1 2 ( a ) f is strictly increasing if f ' x > 0 i.e. if 1 - x 2 x 2 + 1 2 > 0 i.e. if 1 - x 2 > 0 . . . . . . . ∴ x 2 + 1 2 > 0 i.e. if 1 > x 2 i.e. if x 2 < 1 i.e. if - 1 < x < 1 ∴ f is strictly increasing if - 1 < x < 1 . ( b ) f is strictly decreasing if f ' x < 0 i.e. if 1 - x 2 x 2 + 1 2 < 0 i.e. if 1 - x 2 < 0 . . . . . . . ∴ x 2 + 1 2 > 0 i.e. if 1 < x 2 i.e. if x 2 < 1 i.e. if x > 1 or x < - 1 ∴ f is strictly decreasing if x < - 1 or x > 1 . i.e. x ∈ - ∞ , - 1 ∪ 1 , ∞

Show that f x = 3 x + 1 3 x increasing in 1 3 , 1 and decreasing in 1 9 , 1 3 .

Let f x = 3 x + 1 3 x Differentiating w.r.t. x , we get f ' x = d d x 3 x + 1 3 x f ' x = 3 d d x x + 1 3 d d x x - 1 f ' x = 3 × 1 + 1 3 - 1 x - 2 f ' x = 3 - 1 3 x 2 Now, f is increasing if f ' x > 0 and is decreasing if f ' x < 0 . Let x ∈ 1 3 , 3 Then 1 3 < x < 1 ∴ 1 9 < x 2 < 1 ∴ 1 3 < 3 x 2 < 3 ∴ 3 < 1 3 x 2 < 1 3 ∴ - 3 < - 1 3 x 2 < - 1 3 ∴ 3 - 3 < 3 - 1 3 x 2 < 3 - 1 3 ∴ 0 < f ' x < 8 3 ∴ f ' x < 0 for all x ∈ 1 3 , 1 ∴ f is increasing in the interval 1 3 , 1 . Let x ∈ 1 9 , 1 3 Then 1 9 < x < 1 3 1 81 < x 2 < 1 9 1 27 < 3 x 2 < 1 3 27 < 1 3 x 2 < 3 - 27 < - 1 3 x 2 < - 3 3 - 27 < 3 - 1 3 x 2 < 3 - 3 - 24 < f ' x < 0 ∴ f ' x < 0 for all x ∈ 1 9 , 1 3 ∴ f is decreasing in the interval 1 9 , 1 3 . Hence, it is proved.

Show that y = log 1 + x - 2 x 2 + x , x > - 1 , is an increasing function of x throughout its domain.

For the given function y = log 1 + x - 2 x 2 + x On differentiating both sides w.r.t. x we get d y d x = d d x log ( 1 + x ) - 2 x 2 + x d y d x = d d x log 1 + x - d d x 2 x 2 + x d y d x = 1 1 + x - 2 + x d d x 2 x - 2 x d d x 2 + x 2 + x 2 ∵ d d x a b = b d d x a - a d d x b b 2 d y d x = 1 1 + x - 2 + x 2 - 2 x 2 + x 2 d y d x = 1 1 + x - 4 + 2 x - 2 x 2 + x 2 d y d x = 1 1 + x - 2 2 + x 2 d y d x = 2 + x 2 - 4 1 + x 1 + x 2 + x 2 d y d x = x 2 + 4 x + 4 - 4 - 4 x 1 + x 2 + x 2 d y d x = x 2 1 + x 2 + x 2 d y d x = x 2 2 + x 2 1 1 + x For y to be increasing d y d x > 0 , Then x 2 2 + x 2 1 1 + x > 0 If 1 + x > 0 ⇒ x > - 1 ∵ x 2 + x 2 is always + v e . Then, f ' x > 0 Hence, The function is increasing function of x throughout its domain.

The surface area of a spherical balloon is increasing at the rate of 2 cm 2 / sec . At what rate the volume of the balloon is increasing when the radius of the balloon is 6 cm .

Let S be the surface area of a spherical balloon and r is the radius of the balloon. Given, d S d t = 2 cm 2 / sec and r = 6 cm . We know that S = 4 πr 2 Differentiating both sides with respect to time t , we get: d S d t = 8 πr · d r d t ⇒ d S d t = 8 π × 6 × d r d t ⇒ 2 = 48 π · d r d t ⇒ d r d t = 1 24 π cm / sec We know that V = 4 3 πr 3 Differentiating both sides with respect to time t , we get: d V d t = 4 3 π × 3 r 2 × d r d t = 4 π × 36 × 1 24 π ∵ r = 6 cm = 6 cm 3 / sec Hence, the volume of the balloon is increasing at the rate of 6 cm 3 / sec .