Q. 80 - 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 $f$.

$f\left(x\right)=\ln \left(\right(x-1\left)\right(x-2\left)\right)$

Step-by-Step Solution

Answer

The function $f$ is defined everywhere except at $x=1,2$, and the roof of the function is at $x=\frac{3\pm \sqrt{5}}{2}$. The function is positive everywhere except on $(0.36,2.6)$ and it doesn't have a local extrema. The function is increasing on $(2,\infty )$ and decreasing on $(-\infty ,1)$. $\underset{x\to -\infty }{\lim }f\left(x\right)=\infty .\underset{x\to \infty }{\lim }f\left(x\right)=\infty$.

Step 1. Given data

We have the given function,$f\left(x\right)=\ln \left(\right(x-1\left)\right(x-2\left)\right)$

that is,$y=\ln \left(\right(x-1\left)\right(x-2\left)\right)$

Step 2. Graph

The point table of the function is given by,

x	y	(x,y)
0	0.6931	(0,0.6931)
-1	1.7915	(-1,1.7915)
3	0.6931	(3,0.6931)
4	1.7915	(4,1.7915)
5	2.485	(5,2.485)
6	3	(6,3)

The graph of the function is, 

Step 3. Critical points

To find the critical points, let $f\text{'}\left(x\right)=0$

$\begin{array}{c}\frac{d}{dx}(\ln (x-1\left)\right(x-2\left)\right)=0\\ \frac{1}{(x-1)(x-2)}\frac{d}{dx}\left(\right(x-1\left)\right(x-2\left)\right)=0\\ \frac{1}{(x-1)(x-2)}\frac{d}{dx}(x^2-2x-x+2)=0\\ \frac{1}{(x-1)(x-2)}\frac{d}{dx}(x^2-3x+2)=0\\ 2x-3=0\\ \frac{1}{(x-1)(x-2)}(2x-3)=0\\ x=\frac{3}{2}\end{array}$

Hence, $f$ has a critical point are at$x=\frac{3}{2}$ . There are no local extrema. 

Step 4. Sign chart

The sign chart of the function is, 

Step 5. Roots of the function

For the roots of the functions be, 

$\begin{array}{rl}\ln (x-1)(x-2)& =0\\ \ln (x-1)(x-2)& =\ln 1\\ (x-1)(x-2)& =1\\ x^2-x-2x+2& =1\\ x^2-3x+1& =0\\ x& =\frac{-(-3)\pm \sqrt{(-3{)}^2-4\cdot 1\cdot 1}}{2\cdot 1}\\ x& =\frac{3\pm \sqrt{9-4}}{2}\\ x& =\frac{3\pm \sqrt{5}}{2}\end{array}$

Hence the function has a root $x=\frac{3\pm \sqrt{5}}{2}$

Again,

$\begin{array}{rl}\underset{x\to \mathrm{\infty }}{lim}f\left(x\right)& =\underset{x\to \mathrm{\infty }}{lim}[\ln (x-1\left)\right(x-2\left)\right]\\ & =\mathrm{\infty }\\ \underset{x\to -\mathrm{\infty }}{lim}f\left(x\right)& =\underset{x\to -\mathrm{\infty }}{lim}[\ln (x-1\left)\right(x-2\left)\right]\\ & =\mathrm{\infty }\end{array}$

The function $f$ is defined everywhere except at$x=1,2$ , and the roof of the function is at $x=\frac{3\pm \sqrt{5}}{2}$. The function is positive everywhere except on $(0.36,2.6)$ and it doesn't have a local extrema. The function is increasing on $(2,\infty )$ and decreasing on $\backslash \left(\right(-\backslash infty, 1)$ . $\underset{x\to -\infty }{\lim }f\left(x\right)=\infty .\underset{x\to \infty }{\lim }f\left(x\right)=\infty$.