Question

For the function $f$ given by
$f\left(x\right)=3x^4+4x^3-12x^2+12$

• A. Local minima is at $x=1$
• B. Local minima is at $x=0$
• C. Local maxima of $f=12$
• D. Local minima of $f=20$

Answer 1

Answer: (A), (C)

Local maxima of $f=12$

Given, $f\left(x\right)=3x^4+4x^3-12x^2+12$
$\Rightarrow f^{\prime }\left(x\right)=12x^3+12x^2-24x=12x(x-1)(x+2)$
$f^{\prime }\left(x\right)=0$ at $x=0,1$ and $-2$
$\Rightarrow f^{\prime \prime }\left(x\right)=36x^2+24x-24=12(3x^2+2x-2)$
$\Rightarrow f^{\prime \prime }\left(0\right)=-24<0$
$\Rightarrow f^{\prime \prime }\left(1\right)=36>0$
$\Rightarrow f^{\prime \prime }(-2)=72>0$
$\therefore$ By second deriavtive test, $x=0$ is a point of local maxima and local maximum value of $f$ at $x=0$ is $12$.
While $x=1$ and $x=-2$ are the points of local minimum,
$\Rightarrow f\left(1\right)=7$ and $f(-2)=-20$