Write the maximum value of f x = x + 1 x , x < 0 .

Given : f ( x ) = x + 1 x ⇒ f ' ( x ) = 1 - 1 x 2 For a local maxima or a local minima, we must have f ' ( x ) = 0 ⇒ 1 - 1 x 2 = 0 ⇒ x 2 = 1 ⇒ x = 1 , - 1 But x < 0 ⇒ x = - 1 Now, f ' ' ( x ) = 2 x 3 At x = - 1 : f ' ' ( - 1 ) = 2 ( - 1 ) 3 = - 2 < 0 So, x = - 1 is a point of local maximum. Thus, the local maximum value is given by f ( - 1 ) = - 1 + 1 - 1 = - 1 - 1 = - 2

Write the maximum value of f x = x + 1 x , x < 0 .

Given : f ( x ) = x + 1 x ⇒ f ' ( x ) = 1 - 1 x 2 For a local maxima or a local minima, we must have f ' ( x ) = 0 ⇒ 1 - 1 x 2 = 0 ⇒ x 2 = 1 ⇒ x = 1 , - 1 But x < 0 ⇒ x = - 1 Now, f ' ' ( x ) = 2 x 3 At x = - 1 : f ' ' ( - 1 ) = 2 ( - 1 ) 3 = - 2 < 0 So, x = - 1 is a point of local maximum. Thus, the local maximum value is given by f ( - 1 ) = - 1 + 1 - 1 = - 1 - 1 = - 2

Write the minimum value of f ( x ) = x + 1 x , x > 0 .

Given : f ( x ) = x + 1 x ⇒ f ' ( x ) = d d x x + 1 x = 1 - 1 x 2 For a local maxima or a local minima, we must have f ' ( x ) = 0 ⇒ 1 - 1 x 2 = 0 ⇒ x 2 = 1 ⇒ x = 1 , - 1 But x > 0 ⇒ x = 1 Now, f ' ' ( x ) = d d x 1 - 1 x 2 = 2 x 3 At x = 1 , f ' ' ( 1 ) = 2 ( 1 ) 3 = 2 > 0 So, x = 1 is a point of local minimum. Thus, the local minimum value is given by f ( 1 ) = 1 + 1 1 = 1 + 1 = 2

Find the least value of f ( x ) = a x + b x , where a > 0 , b > 0 and x > 0 .

We have, f ( x ) = a x + b x ⇒ f ' ( x ) = d d x a x + b x = a - b x 2 For local maxima or local minima, we must have f ' ( x ) = 0 ⇒ a - b x 2 = 0 ⇒ x 2 = b a ⇒ x = ± b a , But, x > 0 ⇒ x = b a Now, f ' ' ( x ) = d d x a - b x 2 = 2 b x 3 At x = b a f ' ' b a = 2 b b a 3 = 2 a 3 2 b 1 2 > 0 [ ∵ a > 0 and b > 0 ] So, x = b a is a point of local minimum. Hence, the least value is f b a = a b a + b b a = a b + a b = 2 a b

Write the maximum value of f x = x 1 x .

Given : f ( x ) = x 1 x Taking log on both sides, we get log e f ( x ) = 1 x log e x Differentiating with respect to x , we get 1 f ( x ) f ' ( x ) = - 1 x 2 log e x + 1 x 2 ⇒ f ' ( x ) = f ( x ) 1 x 2 ( 1 - log e x ) ⇒ f ' ( x ) = x 1 x 1 x 2 - 1 x 2 log e x ⇒ f ' ( x ) = x 1 x - 2 ( 1 - log e x ) For local maxima or local minima, we must have f ' ( x ) = 0 ⇒ x 1 x - 2 ( 1 - log x ) = 0 ⇒ log x = 1 ⇒ x = e Now, f ' ' ( x ) = d d x x 1 x 1 x 2 - 1 x 2 log e x = x 1 x 1 x 2 - 1 x 2 log x 2 + x 1 x - 2 x 3 + 2 x 3 log x - 1 x 3 ⇒ f ' ' ( x ) = x 1 x 1 x 2 - 1 x 2 log x 2 + x 1 x - 3 x 3 + 2 x 3 log x At x = e f ' ' ( e ) = e 1 e 1 e 2 - 1 e 2 log e 2 + e 1 e - 3 e 3 + 2 e 3 log e = - e 1 e 1 e 3 < 0 So, x = e is a point of local maximum. Thus, the maximum value is given by f ( e ) = e 1 e