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Could someone solve it for me? At least I could see it and try to understand how it has to be done. I really have no clue on how to solve this...

Hello SteamKing, Thank you for your reply! I translated it from Dutch... sorry for my mistake. Indeed, I am trying to check if a function f(x) has a derivative at the gives values of x. For -π/2, I choose f(x) = (sin (x)) and plug-in -π/2. It becomes >> f'(-π/2) = cos (-π/2) = 0,99 = 1 For 0...

The function f: R → R is: f(x) = (tan x) / (1 + ³√x) ; for x ≥ 0, sin x ; for (-π/2) ≤ x < 0, x + (π/2) ; for x < -π/2 _ For -π/2 I would say it is not differentiable since both π and 2 are constants and you can not vary a constant. For 0, I would derivate it by using the first function...

Hello, everybody! The function g: R → R is differentiable in -1, g' (-1) = 2 and g (-1) = -2. Can someone show me (how to argue) that lim g(x), with x → -1, = -2? Thanks in advance!

The function f: R → R is: f(x) = (tan x) / (1 + ³√x) ; for x ≥ 0, sin x ; for (-π/2) ≤ x < 0, x + (π/2) ; for x < -π/2 _ For the interval (0,∞), we are interested in f such that f(x) = (tan x) / (1 + ³√x) ; for x ≥ 0 f(x) = tan x / (1 + x¹ʹ³)            (1 + x¹ʹ³)•sec²x −...