Find the points of local maxima or minima of f x = 1 x 2 + 2 using the first derivative test. Also, find the maximum and minimum values.

Given that f x = 1 x 2 + 2 Taking differentiation with respect to x we get f ' x = x 2 + 2 0 - 2 x 1 x 2 + 2 2 ∵ d d x u v = u d v d x - v d u d x v 2 = - 2 x x 2 + 2 2 For critical points f ' x = 0 ⇒ - 2 x = 0 ⇒ x = 0 ∴ x = 0 is critical point. Now, for x < 0 , f ' x > 0 , and for x > 0 , f ' x < 0 At x = 0 , f ' x changes sign from positive to negative. So, x = 0 is point of maxima. And, Maximum value is f 0 = 1 2

Find the points of local maxima or minima of f x = x 1 - x , x > 0 using the first derivation test. Also, find the local maximum or local minimum values.

Given that f x = x 1 - x , x > 0 Taking differentiation with respect to x we get f ' x = x 2 1 - x - 1 + 1 - x [Since d d x u v = u d v d x + v d u d x ] = - x + 1 - x × 2 2 1 - x = - x + 2 - 2 x 2 1 - x = 2 - 3 x 2 1 - x For Critical Points, we have f ' x = 0 ⇒ 2 - 3 x 2 1 - x = 0 ⇒ 2 - 3 x = 0 ⇒ x = 2 3 Therefore, x = 2 3 is critical point. Now, f ' x = 3 2 3 - x 2 1 - x Now, for x < 2 3 , f ' x > 0 , and for x > 2 3 , f ' x < 0 Thus, f ' x changes sign from positive to negative. ∴ x = 2 3 is point of local maxima. And, Maximum value of f x is f 2 3 = 2 3 1 - 2 3 = 2 3 × 1 3 = 2 3 3 = 2 3 9 .

Find the points of local maxima or local minima of f x = x 2 + 2 x , x > 0 using the first derivative test. Also, find the local maximum or minimum values.

Given that f x = x 2 + 2 x Taking differentiation with respect to x , we get f ' x = 1 2 - 2 x 2 [Since d d x x n = n x n - 1 , n ∈ R ] For critical points f ' x = 0 ⇒ 1 2 - 2 x 2 = 0 ⇒ x 2 - 4 2 x 2 = 0 ⇒ x 2 - 4 = 0 ⇒ x 2 = 4 ⇒ x = ± 2 Since f ' x = x - 2 x + 2 2 x 2 Now, for x < 2 , we have f ' x < 0 , and for x > 2 , we have f ' x > 0 Thus, at x = 2 , f ' x changes sign from negative to positive. ∴ x = 2 is point of minima. And, Minimum value is f 2 = 1 + 1 = 2 . Now, for x < - 2 , f ' x > 0 , and for x > - 2 , f ' x < 0 Thus, at x = - 2 , f ' x changes sign from positive to negative. ∴ x = - 2 is a point of maxima And, the maximum value of f x is f - 2 = - 2 2 + 2 - 2 = - 1 - 1 = - 2