202009250041 Exercise 5.8.1

Find out if any the asymptote(s) of the curve given by:

$y=\displaystyle{\frac{x^2-6x+3}{x-3}}$ .

Attempts.

$\begin{aligned} & & x & -3 & & \\\cline{3-5} x & -3 \Big) & x^2 & -6x & +3 & \\ & & x^2 & -3x & & \\ \cline{3-5} & & & -3x & +3 \\ & & & -3x & +9 \\\cline{4-5} & & & & -6 \end{aligned}$

The curve is given by

$\begin{aligned} y & =\displaystyle{\frac{x^2-6x+3}{x-3}} \\ & = x-3 - \displaystyle{\frac{6}{x-3}} \end{aligned}$ .

Ans. Of the given curve, $x=3$ is a vertical asymptote and $y=x-3$ an oblique asymptote.

Retrospectively, for $f(x)=\displaystyle{\frac{x^2-6x+3}{x-3}}$ , maybe one could also have checked the conditions as follow:

i. $\displaystyle{\lim_{x\to 0}}f(x)$ ;

ii. $\displaystyle{\lim_{x\to 3^{+}}}f(x)$ ;

iii. $\displaystyle{\lim_{x\to 3^{-}}}f(x)$ ;

iv. $\displaystyle{\lim_{x\to +\infty}}f(x)$ ; and

iv. $\displaystyle{\lim_{x\to -\infty}}f(x)$ .

i.

$\begin{aligned} \lim_{x\to 0}f(x) & = \lim_{x\to 0} \frac{x^2-6x+3}{x-3} \\ & = \frac{(0)^2-6(0)+3}{(0)-3} \\ & = -1 \end{aligned}$

iii.

$\begin{aligned} \lim_{x\to 3^{-}}f(x) & = \lim_{x\to 3^{-}} \frac{x^2-6x+3}{x-3} \\ & = \lim_{x\to 3^{-}} \bigg( x-3-\frac{6}{x-3} \bigg) \\ & = 3^{-} -3-\frac{6}{3^{-}-3} \\ & = (3^{-}-3) + \bigg( \frac{6}{3-3^{-}} \bigg) \\ & \\ \hline & \\ & \because 3^{-} < 3 \\ & \therefore 3^{-}-3<0\qquad \textrm{and}\qquad 3-3^{-}>0 \\ & \dots \textrm{excuse me for writing with no more ado,}\dots \\ & \\ \hline & \\ & = 0^{-} + \frac{6}{0^{+}} \\ & = +\infty \\ \end{aligned}$

ii., iv., and iv. are not checked.

Just work with it like whom I was taught:

First, find the slope of tangent to the curve,

$\begin{aligned} f(x) & = \frac{x^2-6x+3}{x-3} \\ f'(x) & = \frac{\mathrm{d}}{\mathrm{d}x}\big( f(x) \big) \\ & = \frac{(x-3)(2x-6)-(x^2-6x+3)(1)}{(x-3)^2} \\ & = \frac{(2x^2-12x+18)-(x^2-6x+3)}{(x-3)^2} \\ & = \frac{x^2-6x+15}{x^2-6x+9} \\ & = 1 + \frac{6}{x^2-6x+9} \end{aligned}$

The purpose is to find out any crests or troughs, because at the tip or bottom of a plotted curve, the slope of tangent will be zero, i.e., $f'(x)=0$ .

Secondly, find the change of slope of tangent to the curve,

$\begin{aligned} f''(x) & = \frac{\mathrm{d}}{\mathrm{d}x}\big[ f'(x) \big] \\ & = \frac{\mathrm{d}}{\mathrm{d}x} \bigg( 1+\frac{6}{x^2-6x+9} \bigg) \\ & = \frac{\mathrm{d}}{\mathrm{d}x} \big( 1 \big) + \frac{\mathrm{d}}{\mathrm{d}x} \bigg( \frac{6}{x^2-6x+9} \bigg) \\ & = 0 + \frac{(x^2-6x+9)(0)+(6)(2x-6)}{(x^2-6x+9)^2} \\ & = \frac{12x-36}{(x^2-6x+9)^2} \\ \end{aligned}$

The purpose is to find out if there were any inflexion points (i.e., $f''(x)=0$ ) upon which curvature changes sign.

It begins, assuming $x\neq 3$ , I will set $f(x)=0$ :

$\begin{aligned} f(x) & = 0 \\ \Leftrightarrow \qquad\qquad x^2-6x+3 & = 0 \\ x & = \frac{-(-6)\pm \sqrt{(-6)^2-4(1)(3)}}{2(1)} \\ & = 3 \pm \sqrt{6} \\ \end{aligned}$

The $x$ -intercepts are thus two, $3+\sqrt{6}$ and $3-\sqrt{6}$ .

Then I needs set $f'(x)=0$ with the same assumption $x\neq 3$ .

$\begin{aligned} f'(x) & = 0 \\ \Leftrightarrow \qquad \qquad 1+\frac{6}{x^2-6x+9} & = 0 \\ & \\ \textrm{\dots \quad by assumption }x\neq 3 & \enspace \textrm{jump legitimately to the next step\quad \dots} \\ & \\ x^2-6x+9+6 & = 0 \\ x^2 -6x +15 & = 0 \\ x & = \frac{-(-6)\pm \sqrt{(-6)^2-4(1)(15)}}{2(1)} \\ & = 3 \pm \sqrt{6}\, \textrm{i} \end{aligned}$

As curvature here permits of no nonreal points, there are neither convex point(s) nor concave point(s).

Besides, there are none any one point of inflexion because if on one hand $f''(x)=0$ and on the other $x\neq 3$ be assumed,

$\begin{aligned} \frac{12x-36}{(x^2-6x+9)^2} & = 0 \\ \Rightarrow 12x-36 & = 0 \\ \Rightarrow x & =3\qquad\qquad\qquad \bot \\ \end{aligned}$

it would have resulted in contradiction.

In conclusion, this exercise was overdone.

M	T	W	T	F	S	S
	1	2	3	4	5	6
7	8	9	10	11	12	13
14	15	16	17	18	19	20
21	22	23	24	25	26	27
28	29	30

Leave a Reply Cancel reply