Question 1194169
y=x^2+(2x-1)^2
y'=2x+2(2x-1)*2=2x+8x-4=10x-4
set that equal to 0
10x=4
x=0.4
2nd derivative is 10 and is always positive so this is a minimum
alternative way into expand function to x^2+4x^2-4x+1=5x^2-4x-1
find vertex by x=-b/2a=-4/-10 or +0.4
{{{graph(300,300,-1,2,-10,10,x^2+(2x-1)^2)}}}
--
y'=2 pi-2(x-2)=2pi-2x+4
set equal to 0
2x=2pi+4
x=2+pi or 5.28
y''=-2, and this will be a maximum.
{{{graph(300,300,-10,10,-10,30,2*pi*x-(x-2)^2)}}}