Q. 62 - 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 $f$.

$f\left(x\right)=x\left(x^2-4\right)$.

Step-by-Step Solution

Answer

The graph for the function $f\left(x\right)=x\left(x^2-4\right)$ is,

Step 1 . Given information

$f\left(x\right)=x\left(x^2-4\right)$.

Step 2 . Let y = x x 2 - 4 .

Now point table for the function which is given by,

$x$	$y$	$\left(x,y\right)$
$0$	$0$	$\left(0,0\right)$
$1$	$-3$	$\left(1,-3\right)$
$-1$	$3$	$\left(-1,3\right)$
$2$	$0$	$\left(2,0\right)$
$-2$	$0$	$\left(-2,0\right)$

Step 3 . The graph for the function is,

Step 4 . Now for critical point, f ' x = 0 .

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

Therefore, $f$ has a critical point at $x=\pm \frac{2}{\sqrt{3}}$.But, that is a local extremum at $x=-\frac{2}{\sqrt{3}}$ and $x=\frac{2}{\sqrt{3}}$.

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

Therefore, the function $f$ is increasing on $x=-\frac{2}{\sqrt{3}}$ and on $x=\frac{2}{\sqrt{3}}$which may also written as, $\left(-\infty ,-\frac{2}{\sqrt{3}}\right)\cup \left(\frac{2}{\sqrt{3}},\infty \right)$ and $f$ is decreasing on $\left(-\frac{2}{\sqrt{3}},\frac{2}{\sqrt{3}}\right)$.

Again,

$$\begin{gathered} \underset{x\to -\infty }{\lim }f\left(x\right)=\underset{x\to -\infty }{\lim }\left(x\left(x^2-4\right)\right) \\ =-\infty \\ \underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }\left(x\left(x^2-4\right)\right) \\ =\infty \end{gathered}$$

Therefore, the function is defined everywhere, roots at $x=-2,0,2$. Critical point at $x=\pm \frac{2}{\sqrt{3}}$. that is a local minimum at $x=-\frac{2}{\sqrt{3}}$  and local maximum at $x=\frac{2}{\sqrt{3}}$.

The function $f$ is increasing on $\left(-\frac{2}{\sqrt{3}},\frac{2}{\sqrt{3}}\right)$ including $x=2$. And the limits are, $\underset{x\to -\infty }{\lim }f\left(x\right)=-\infty ; \underset{x\to \infty }{\lim }f\left(x\right)=\infty$.