Question

For the function f(x)=x³-6x²-12 determine

a) intervals of increase and decrease

b) location of max and minimum points

c) interval of concavity down or up

d) locator of points of inflection

Answer 1

To solve the items find previously first and second derivatives of given function

$f(x)=x^3-6x^2-12$

$f'(x)=3x^2-6\cdot2x-0=3x^2-12x$

$f''(x)=3\cdot2x-12=6x-12$

a) intervals of increase and decrease

Find intervals of increase solving the inequality

$f'(x)>0$

$3x^2-12x>0$

$3x(x-4)>0$

$\left\{\begin{array}{l}3>0\\ x(x-4)>0\end{array}\right.\Rightarrow$ $x(x-4)>0\Rightarrow$

$\left[\begin{array}{l}\left\{\begin{array}{l}x>0\\x-4>0\end{array}\right.\\ \\ \left\{\begin{array}{l}x<0\\x-4<0\end{array}\right.\end{array}\right.\Rightarrow$ $\left[\begin{array}{l}\left\{\begin{array}{l}x>0\\x>4\end{array}\right.\\ \\ \left\{\begin{array}{l}x<0\\x<4\end{array}\right.\end{array}\right.\Rightarrow$ $\left[\begin{array}{l}x>4 \\ x<0\end{array}\right.\Rightarrow$ $x\in(-\infty;0)\cup(4;+\infty)$

Find intervals of decrease solving the inequality

$f'(x)<0$

$3x^2-12x<0$

$3x(x-4)<0$

$\left\{\begin{array}{l}3>0\\ x(x-4)<0\end{array}\right.\Rightarrow$ $x(x-4)<0\Rightarrow$

$\left[\begin{array}{l}\left\{\begin{array}{l}x>0\\x-4<0\end{array}\right.\\ \\ \left\{\begin{array}{l}x<0\\x-4>0\end{array}\right.\end{array}\right.\Rightarrow$ $\left[\begin{array}{l}\left\{\begin{array}{l}x>0\\x<4\end{array}\right.\\ \\ \left\{\begin{array}{l}x<0\\x>4\end{array}\right.\end{array}\right.\Rightarrow$ $\left[\begin{array}{l}0<x<4 \\ x\in\emptyset\end{array}\right.\Rightarrow$ $x\in(0;4)$

b) location of maximum and minimum points

Find critical points solving the equation

$f'(x)=0$

$3x^2-12x=0$

$3x(x-4)=0$

$\left[\begin{array}{l}x=0 \\ x=4\end{array}\right.$

At the point $x_1=0$ the derivative swithes its sign from $+$ to $-$ . Hence the function switches from increasing to decreasing at the point. Thus $x_1=0$ is the maximum point.

$f(0)=12$

At the point $x_2=4$ the derivative swithes its sign from $-$ to $+$ . Hence the function switches from decreasing to increasing at the point. Thus $x_2=4$ is the minimum point.

$f(4)=4^3-6\cdot4^2-12=64-96-12=-44$

c) intervals of concavity down or up

Find intervals of concavity down solving the inequality

$f''(x)>0$

$6x-12>0$

$6(x-2)>0$

$\left\{\begin{array}{l}6>0\\ x-2>0\end{array}\right.\Rightarrow$ $x>2\Rightarrow$ $x\in(2;+\infty)$

Find intervals of concavity up solving the inequality

$f''(x)<0$

$6x-12<0$

$6(x-2)<0$

$\left\{\begin{array}{l}6>0\\ x-2<0\end{array}\right.\Rightarrow$ $x<2\Rightarrow$ $x\in(-\infty;2)$

d) locator of points of inflection

Find points of inflection solving the equation

$f''(x)=0$

$6x-12=0$

$6(x-2)=0$

$x=2$

At the point $x=2$ the second derivative swithes its sign from $-$ to $+$ . Hence the graph of function switches from concavity up to concavity down at the point. Thus $x=2$ is the inflection point.

$f(2)=2^3-6\cdot2^2-12=8-24-12=-28$