Question

Can someone pls help me with proving that (f)x^2-2|x| is increasing over [1;infinity] and decreasing over [0;1] then deduce that (f) admits a minimum to be determined

Answer 1

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             

Answer 2

Derivatives galore. Don't forget you might need to split the function because of the absolute value.