here is my answer..confirm if its rite or not
g(x)= f(x+1) x+1<2
-4 1<x<2
f(x) x≥2
f(x)=x2-4x
g(x)=min f(t) ; x≤t≤x+1 ; 0≤x<4
i know this type ahs been discussed before...but i just need graph to check myself....no calculations....[1]
here is my answer..confirm if its rite or not
g(x)= f(x+1) x+1<2
-4 1<x<2
f(x) x≥2
yes akari, your graph is not correct. There is an interval when g(x) is a constant.
solving f(x) i am getting (y+4)=(x-2)2
so graph of f(x) is an upward opening parabola with vertex at (2,-4)
next how to proceed?
how to find g(x)