Let f(x)=2x3−3x2−12x+5 ∴f′(x)=6x2−6x−12 Put f′(x)=0, for maxima or minima. ∴6x2−6x−12=0 ⇒x2−x−2=0 ⇒x2−2x+x−2=0 ⇒(x−2)(x+1)=0 ⇒x=−1,2 Now, f"(x)=12x−6 f"(−1)=−12−6=−18<0 ∴f(x) is maximum at x=−1. But x=4 f(x)=37 ∴ The largest value of f(x)