Question

Find the points of local maximum and local minimum of the function$f\left(x\right)=3x^3+6x^2-12x+12$

Answer 1

Consider the given equation.

$f\left(x\right)=3x^3+6x^2-12x+12$

Differentiate $f\left(x\right)$ with respect to $x$.

$f^{\prime }\left(x\right)=9x^2+12x-12$

We know that, at local maxima and local minima,

$f^{\prime }\left(x\right)=0$

$9x^2+12x-12=0$

$3x^2+4x-4=0$

$3x^2+6x-2x-4=0$

$3x(x+2)-2(x+2)=0$

$(x+2)(3x-2)=0$

$\Rightarrow x=-2,\frac{2}{3}$

Now,

$f^{\prime \prime }\left(x\right)=18x+12$

Here,

${f^{\prime \prime }\left(x\right)]}_{x=-2}=18\times (-2)+12=-36+12=-24<0$

Therefore, $x=-2$ is the point of local maxima.

Again,

${f^{\prime \prime }\left(x\right)]}_{x=\frac{2}{3}}=18\times \left(\frac{2}{3}\right)+12=12+12=24>0$

Therefore, $x=\frac{2}{3}$ is the point of local minima.