f(x)=x2+2x+4x2−3x−6​f′(x)=(x2+2x+4)2(x2+2x+4)(2x−3)−(x2−3x−6)(2x+2)​f′(x)=(x2+2x+4)25x(x+4)​​Sign scheme for f′(x)Here, f is decreasing in the interval (−2,−1) and f is increasing in the interval (1,2).Now, f(−4)=611​,f(0)=2−3​[from Eq. (i)] and limx→±∞​f(x)=1∴ Range =[2−3​,611​]Hence, f(x) is into.f(x) has local maxima at x=−4and local minima at x=0.