Q. 71 - 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 $f$.

$f\left(x\right)=\frac{{\left(x-1\right)}^2}{x^2+x-6}$.

Step-by-Step Solution

Answer

The graph for the function $f\left(x\right)=\frac{{\left(x-1\right)}^2}{x^2+x-6}$ is,

Step 1 . Given information

$f\left(x\right)=\frac{{\left(x-1\right)}^2}{x^2+x-6}$.

Step 2 . Let y = x - 1 2 x 2 + x - 6 .

Now point table for the function is given by,

$x$	$y$	$\left(x,y\right)$
$-4$	$\frac{25}{6}$	$\left(-4,\frac{25}{6}\right)$
$-6$	$\frac{49}{24}$	$\left(-6,\frac{49}{24}\right)$
$0$	$-\frac{1}{6}$	$\left(0,-\frac{1}{6}\right)$
$1$	$0$	$\left(1,0\right)$
$3$	$\frac{2}{3}$	$\left(3,\frac{2}{3}\right)$

Step 3 . The graph for the function is,

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

$$\begin{gathered} \frac{d}{dx}\frac{{\left(x-1\right)}^2}{x^2+x-6}=0 \\ \frac{(x^2+x-6)2\left(x-1\right)-{\left(x-1\right)}^2\left(2x+1\right)}{{\left(x^2+x-6\right)}^2}=0 \\ (x^2+x-6)2\left(x-1\right)-{\left(x-1\right)}^2\left(2x+1\right)=0 \\ \left(x-1\right)\left[2x^2+2x-12-2x^2-x+2x+1\right]=0 \\ \left(x-1\right)\left(3x-11\right)=0 \\ x=1,\frac{11}{3} \end{gathered}$$

Therefore, $f$ has two critical points at $x=1,\frac{11}{3}$.So, it has no local extrema.

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

For roots of the function,

$$\begin{gathered} \frac{{\left(x-1\right)}^2}{x^2+x-6}=0 \\ {\left(x-1\right)}^2=0 \\ x-1=0 \\ x=1 \end{gathered}$$

Therefore, $x$ has a root which is $1$.

Step 6 . Therefore, the function f is defined everywhere except at x = - 3 and x = 2 where there are a vertical asymptotes.

Again,

$$\begin{gathered} \underset{x\to -\infty }{\lim }f\left(x\right)=\underset{x\to -\infty }{\lim }\frac{{\left(x-1\right)}^2}{x^2+x-6} \\ =\infty \\ \underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }\frac{{\left(x-1\right)}^2}{x^2+x-6} \\ =\infty \end{gathered}$$