If f x = x + 1 x - 1 , show that f f x = x .

Given, f x = x + 1 x - 1 ∴ f f x = f x + 1 f x - 1 = x + 1 x - 1 + 1 x + 1 x - 1 - 1 = x + 1 + x - 1 x + 1 - x + 1 = 2 x 2 = x Hence, proved that f f x = x .

If f x = x + 1 x - 1 , show that f f x = x .

Given, f x = x + 1 x - 1 ∴ f f x = f x + 1 f x - 1 = x + 1 x - 1 + 1 x + 1 x - 1 - 1 = x + 1 + x - 1 x + 1 - x + 1 = 2 x 2 = x Hence, proved that f f x = x .

If f x = x 2 , when x < 0 x , when 0 ≤ x < 1 1 x , when x > 1 Find f 1 2 , f - 2 , f 1 , f 3 , f - 3 .

Given f x = x 2 , when x < 0 x , when 0 ≤ x < 1 1 x , when x > 1 Now, for f 1 2 , 0 ≤ x < 1 ⇒ f ( x ) = x So, f 1 2 = 1 2 when x < 0 , f ( x ) = x 2 So, f - 2 = - 2 2 = 4 ∵ f x = 1 x for x ≥ 1 So, f 1 = 1 1 = 1 ∵ f x = 1 x for x > 1 So, f 3 = 1 3 ∵ - 3 is imaginary, it does not exist in given intervals. ∴ f ( - 3 ) given is not defined or does not exist.

If f x = x 3 - 1 x 3 , show that f x + f 1 x = 0 .

Given f x = x 3 - 1 x 3 ∴ f 1 x = 1 x 3 - 1 1 x 3 = 1 x 3 - x 3 ∴ f x + f 1 x = x 3 - 1 x 3 + 1 x 3 - x 3 = 0 Hence, proved.

If f x = 2 x 1 + x 2 , show that f tan ⁡ θ = sin ⁡ 2 θ

Given f x = 2 x 1 + x 2 Replacing x with tan θ in the function ∴ f t a n θ = 2 t a n θ 1 + tan 2 θ = sin 2 θ ∵ sin 2 θ = 2 tan θ 1 + tan 2 θ Hence, proved that f tanθ = sin 2 θ .

If f x = x - 1 x + 1 , then show that f 1 x = - f x .

Given, f x = x - 1 x + 1 ∴ f 1 x = 1 x - 1 1 x + 1 = 1 - x 1 + x = - x - 1 x + 1 = - f x Hence, proved that f 1 x = - f x .

If f x = x 2 , when x < 0 x , when 0 ≤ x < 1 1 x , when x > 1 , then find f - 2 .

Given that, f x = x 2 , when x < 0 x , when 0 ≤ x < 1 1 x , when x > 1 To find, f - 2 - 2 < 0 Therefore, - 2 is substituted in f x = x 2 Therefore, f - 2 = 4

If f x = x 2 , when x < 0 x , when 0 ≤ x < 1 1 x , when x > 1 , then find f - 3 .

Given that, f x = x 2 , when x < 0 x , when 0 ≤ x < 1 1 x , when x > 1 To find, f - 3 - 3 = 3 i is a complex number. f x is defined only for real values. Therefore, f - 3 = undefined

If f x = x 2 , when x < 0 x , when 0 ≤ x < 1 1 x , when x > 1 , then find f 3 .

Given that, f x = x 2 , when x < 0 x , when 0 ≤ x < 1 1 x , when x > 1 To find, f 3 3 = 1 . 732 > 1 Therefore, 3 is to be substituted in f x = 1 x . Therefore, f 3 = 1 3

If f x = x 2 , when x < 0 x , when 0 ≤ x < 1 1 x , when x > 1 , then find f 1 .

Given that, f x = x 2 , when x < 0 x , when 0 ≤ x < 1 1 x , when x > 1 To find, f 1 f x is discontinuous at x = 1 . f x is defined for values x > 1 and x < 1 , but not for x = 1 . Therefore, f 1 = undefined