Respuesta :
Derivatives galore. Don't forget you might need to split the function because of the absolute value.
hello :
f(x) = x²-2x if x ≥ 0 ( |x | = x)
f'(x) = 2x-2
f'(x) = 0 ... x=1
f'(x) < 0 .... x in : ]0 ; 1[
f'(x) > 0 ..... x in : ]1 ; + ∞[
f(1) is a minimum : f(1) = 1² -2(1)= -1
f(x) = x²-2x if x ≥ 0 ( |x | = x)
f'(x) = 2x-2
f'(x) = 0 ... x=1
f'(x) < 0 .... x in : ]0 ; 1[
f'(x) > 0 ..... x in : ]1 ; + ∞[
f(1) is a minimum : f(1) = 1² -2(1)= -1
