Q. 61 - Calculus

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 $f$.

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

Step-by-Step Solution

Answer

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

Step 1 . Given information

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

Step 2 . Let y = x 3 - 6 x 2 + 12 x .

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 $\underset{x\to \infty }{\lim }f\left(x\right)=\infty$.