The draw is:
Now, to find the extreme points we need to calculate the derivative and solve it for equal zero.
[tex]\begin{gathered} f(x)=3x^4+4x^3−12x^2+12 \\ f^{\prime}(x)=12x^3+12x^2−24x \end{gathered}[/tex]Let's solve = 0:
[tex]\begin{gathered} f^{\prime}(x)=12x^3+12x^2−24x^=0 \\ 12x^3+12x^2−24x=0 \\ 12x(x^2+x-2)=0 \\ x^2+x-2=0 \\ (x+2)(x-1)=0 \\ x=-2 \\ x=1 \\ x=0 \end{gathered}[/tex]We have the extreme points: x= -2, x=1, x=0. To compute what point is minimal or maximal we have to evaluate each point. For x= -2
[tex]\begin{gathered} f(x)=3x^4+4x^3−12x^2+12 \\ f(-2)=3(-2)^4+4(-2)^3−12(-2)^2+12 \\ f(-2)=48-32-48+12=-20 \end{gathered}[/tex]Now for x=1:
[tex]f(1)=3(1)^4+4(1)^3-12*1^2+12=3+4-12+12=7[/tex]Now for x=0:
[tex]f(0)=3*0+4*0-12*0+12=12[/tex]Comparing these 3 points, we have local minimals at -2 and 1, and a local maximal at 0.
