Question

Ex 2.6

8. find the stationary points on the curve y=3x^4 -4x³ -12x² +1 and determine whether they are maximum or minimum turning points

Answer 1

`y=3x^4 -4x^3 -12x^2 +1\\ y'=12x^3-12x^2-24x\\\\ 12x^3-12x^2-24x=0\\ 12x(x^2-x-2)=0\\ 12x(x^2+x-2x-2)=0\\ 12x(x(x+1)-2(x+1))=0\\ 12x(x-2)(x+1)=0\\ x=0 \vee x=2 \vee x=-1\\\\ \forall{x\in(-\infty,-1)\cup(0,2)}\ y'<0\Rightarrow y\searrow\\ \forall{x\in(-1,0)\cup(2,\infty)}\ y'>0\Rightarrow y\\earrow\\ \Downarrow\\ y(0)=1=y_(max)\\ y(2)=3\cdot2^4-4\cdot2^3-12\cdot2^2+1=48-32-48=-32=y_(min)`

`y(-1)=3\cdot(-1)^4-4\cdot(-1)^3-12\cdot(-1)^2+1=3+4-12+1=-4=\\ =y_(min)`