Question

Find local maximum and local minimum values of the function $f$ given by $f\left(x\right)=3x^4+4x^3-12x^2+12$.

Answer 1

$f\left(x\right)=3x^4+4x^3-12x^2+12$
Differentiating w.r.t. $x,$
$f^{\prime }\left(x\right)=12x^3+12x^2-24x$
$\Rightarrow f^{\prime }\left(x\right)=12x(x^2+x-2)$
$\Rightarrow f^{\prime }\left(x\right)=12x(x-1)(x+2)$
Putting $f^{\prime }\left(x\right)=0$
$\Rightarrow 12x(x-1)(x+2)=0$
$\Rightarrow x(x-1)(x+2)=0$
$\Rightarrow x=-2,0,1$
Critical points are $x=-2,0,1$
$f^{\prime }\left(x\right)=12x^3+12x^2-24x$

Differentiation w.r.t. $x$
$f^{\prime \prime }\left(x\right)=36x^2+24x-24$
$\Rightarrow f"\left(x\right)=12(3x^2+2x-2)$
Using second derivative test
At $x=-2$
$f^{\prime \prime }(-2)=12\left(3\right(-2{)}^2+2(-2)-2)$
$=12(12-4-2)$
$=72>0$
$\therefore x=-2$ is point of local minima.
At $x=0$
$f^{\prime \prime }\left(0\right)=12\left(3\right(0{)}^2+2\left(0\right)-2)$
= 12(0+0-2)\)
$=-24<0$

$\therefore x=0$ is point of local maxima
At $x=1$
$f^{\prime \prime }\left(1\right)=12\left[3\right(1)+2(1)-2]$
$=36>0$
$\therefore x=1$ is pont of local minima.
$f\left(x\right)=3x^4+4x^3-12x^2+12$
Local minima points are : $x=-2,1$
Local minimum values are:

$f(-2)=3(-2{)}^4+4(-2{)}^3-12(-2{)}^2+12$
$\Rightarrow f(-2)=48-32-48+12=20$
And
$f\left(1\right)=3+4-12+12=7$
Local maxima point is $x=0$
Local maximum value is $f\left(0\right)=12$