Find the extreme points of the function y = x e - x , x ∈ ( - 2 , 2 ) . Specify if the extremum is a maximum or minimum. Find the extreme values.

Given that, y = x e - x , x ∈ ( - 2 , 2 ) A function f x have a maximum value when f ' x = 0 and f ' ' x < 0 and a maximum value when f ' x = 0 and f ' ' x > 0 . We can also find values of f x at critical points and conclude between maximum and minimum values. Now, d y d x = e - x - x e - x d y d x = 0 ⇒ e - x - x e - x = 0 ⇒ e - x ( 1 - x ) = 0 ⇒ x = 1 ; e - x ≠ 0 as x ∈ ( - 2 , 2 ) d 2 y d x 2 = - e - x - e - x + x e - x At x = 1 , d 2 y d x 2 = - 2 e - 1 + e - 1 = - e < 0 ∴ y is maximum at x = 1 ∴ Maximum value of y = 1 · e - 1 i.e., y = 1 e .

Find the extreme points of the function y = x e - x , x ∈ ( - 2 , 2 ) . Specify if the extremum is a maximum or minimum. Find the extreme values.

Given that, y = x e - x , x ∈ ( - 2 , 2 ) A function f x have a maximum value when f ' x = 0 and f ' ' x < 0 and a maximum value when f ' x = 0 and f ' ' x > 0 . We can also find values of f x at critical points and conclude between maximum and minimum values. Now, d y d x = e - x - x e - x d y d x = 0 ⇒ e - x - x e - x = 0 ⇒ e - x ( 1 - x ) = 0 ⇒ x = 1 ; e - x ≠ 0 as x ∈ ( - 2 , 2 ) d 2 y d x 2 = - e - x - e - x + x e - x At x = 1 , d 2 y d x 2 = - 2 e - 1 + e - 1 = - e < 0 ∴ y is maximum at x = 1 ∴ Maximum value of y = 1 · e - 1 i.e., y = 1 e .

Show that the function ln | x | , x ≠ 0 do not possess maximum or minimum.

Let, f ( x ) = ln | x | For a function f x if f ' ( c ) = f ' ' ( c ) = ⋯ = f n - 1 ( c ) = 0 and f n ( c ) ≠ 0 and n is even, then f n ( c ) > 0 ⇒ x = c is point of local minimum f n ( c ) < 0 ⇒ x = c is point of local maximum If n is odd, then x = c gives neither maximum or minimum. It is a point of inflection on f x curve. Now, f ' x = 1 | x | · d d x | x | = 1 | x | · | x | x = 1 x f ' x = 0 ⇒ 1 x = 0 ⇒ x = ∞ f ' ' x = - 1 x 2 f ' ' ( ∞ ) = 0 f ' ' ' x = 2 x 3 f ' ' ' ( ∞ ) = 0 For any integer value of n , f n ( x ) = 0 To find maximum or minimum f n ( x ) ≠ 0 and n should be even. ∴ f ( x ) = ln | x | do not possess maximum or minimum. Hence, shown.

Use the function f x = x 1 x , x > 0 to show that e n > n e .

Given, f x = x 1 x , x > 0 ∴ log f x = 1 x log x Differentiating w.r.t x , we get 1 f ( x ) × f ' x = 1 x × 1 x + log e x - 1 x 2 ⇒ f ' x = x 1 x x 2 1 - log e x ...(i) Consider the function for x = e to x = n , n > e ∴ For e < x < n , log e x > 1 as x > e then it follows from (i) that f ' ( x ) < 0 for x > e . Hence, f x is decreasing function of x for x > e . ⇒ Since n > e we conclude f n < f e ⇒ n 1 n < e 1 e ⇒ n 1 n e n < e 1 e e n ⇒ n e < e n Hence, shown.

Prove the inequality x 2 e - x 2 ≤ e - 1 , x ∈ R .

Given, x 2 · e - x 2 ≤ e - 1 , x ∈ R Now, e x + e - x = ∑ n x n n ! + ∑ n - 1 n x n n ! ∵ e x = ∑ n x n n ! = 2 ∑ n x n n ! = 2 ∑ n x 2 n ( 2 n ) ! ≤ 2 ∑ n x 2 n n ! , where n is even ≤ 2 e x 2 ∵ ∑ n x 2 n n ! = e x 2 ⇒ e x + e - x ≤ 2 e x 2 ⇒ e - x 2 ≤ 2 e x + e - x ⇒ e - x 2 ≤ 2 1 + x + x 2 2 ! + x 3 3 ! + ⋯ + 1 - x + x 2 2 ! - x 2 3 ! + ⋯ ⇒ e - x 2 ≤ 2 2 + 2 x 2 2 ! + ⋯ - ⇒ e - x 2 ≤ 1 1 + 1 1 ! + ⋯ ⇒ x 2 · e - x 2 ≤ 1 e ∵ e = 1 + 1 1 ! + 1 2 ! + ⋯ ⇒ x 2 · e - x 2 ≤ e - 1 Hence, proved.