If Q = 4v3 + 3v2 , then the value of 'v' such that, there exist maxima of 'Q'

Q=4v3+3v2⇒dQdv=12V2+6V=0⇒v=0,-12Now, d2Qdv2=24v+6For v=0, d2Qdv2=6(+ve)For v=-12 , d2Qdv2=−12+6=−6(−ve)Thus, v=−1/2 for maximum Q