f'(x) = 4x3 - 196x
let f'(x) = 0
4x3 - 196x =0
x(4x2 - 196)=0
x=0,+7,-7
consider:
f"(x) = 12x2 -196
f"(0) = -196 <0 , x = 0 is abs max, and its value ymax = 0-0+9 = 9
f"(7)=f(-7) = 12*49 - 196 = 392 >0
x = -7, 7 is abs min, and their values are
ymin = (7)4 - 98(7)2 + 9 = (-7)4 - 98(-7)2 + 9= -2392