Question

Sketch careful, labeled graphs of each function $f$in Exercises $57-82$by hand, without consulting a calculator or graphing utility. As part of your work, make sign charts for the signs, roots, and undefined points of $f$and $f^{\text{'}}$and examine any relevant limits so that you can describe all key points and behaviors of $f$.

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

Short answer

The graph for the function$f\left(x\right)=x^3-6x^2+12x$ is,

Step-by-step solution

Step 1. Given information

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

Step 2. Let y=x3-6x2+12x.

Now point table for the function is given by,

$x$	$y$	$\left(x,y\right)$
$0$	$0$	$\left(0,0\right)$
$1$	$7$	$\left(1,7\right)$
$2$	$8$	$\left(2,8\right)$
$3$	$9$	$\left(3,9\right)$

Step 3. The graph of the function is shown below:

Step 4. Now for critical point f'x=0.

$$\begin{gathered} \frac{d}{dx}\left(x^3-6x^2+12x\right)=0 \\ 3x^2-12x+12=0 \\ {x}^2-4x+4=0 \\ (x-2{)}^2=0 \\ x-2=0 \\ x=2 \end{gathered}$$

Therefore,$x$has a critical point at$x=2$. But that it is not a local extremum.

Step 5. The sign chart of f is shown below:

Therefore, the function $f$is increasing everywhere even at $x=2$that $f$is increasing on $\left(-\infty ,2\right)\cup \left(2,\infty \right)$.

Again,

$$\begin{gathered} \underset{x\to -\infty }{\lim }f\left(x\right)=\underset{x\to -\infty }{\lim }\left(x^3-6x^2+12x\right) \\ =-\infty \\ \underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }\left(x^3-6x^2+12x\right) \\ =\infty \end{gathered}$$

Therefore, the function is defined everywhere, roots at $x=0$. Critical point $x=2$which is not a local extremum.

The function $f$is increasing on $\left(-\infty ,2\right)\cup \left(2,\infty \right)$including$x=2$and the limits are$\underset{x\to -\infty }{\lim }f\left(x\right)=-\infty$and$\underset{x\to \infty }{\lim }f\left(x\right)=\infty$.