f(x) = x^2 -4x +5

f ' (x) = 2x - 4 which is simply a line yint - 4 and x int 2 with slope 2

The only 0 is 2x-4=0 or x=2

It is fairly easy to see then f ' > 0 if x >2 therefore f is incr

f' < 0 if x < 2 therefore f is decr

Don't know where the -1 < x< 1 comes from but it isn't correct