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Question Number 79883 by MJS last updated on 29/Jan/20

to Sir Jagoll (a n d o f c o u r s e e v e r y b o d y e l s e)

(1)

y = \frac{x^{2} - x - 6}{x^{2} - 3 x - 4} =

= \frac{(x - 3) (x + 2)}{(x - 4) (x + 1)} \Rightarrow

\Rightarrow {\begin{cases} zeros at x = - 2; x = 3 \\ vertical asymptotes at x = - 1; x = 4 \end{cases}

defined for x \in R ∖ {- 1; 4}

range : transforming y = \frac{x^{2} - x - 6}{x^{2} - 3 x - 4} to

x^{2} - \frac{3 y - 1}{y - 1} x - \frac{2 (2 y - 3)}{y - 1} = 0

D = \frac{25 y^{2} - 46 y + 25}{4 {(y - 1)}^{2}} > 0 \forall x \in R \Rightarrow

\Rightarrow {\begin{cases} range = R \\ horizontal asymptote at y = 1 (*) \end{cases}

(*) {doesn}^{'} t mean y = 1 is not within range!!!

x = 1 \Rightarrow y = 1

y^{'} = - \frac{2 (x^{2} - 2 x + 7)}{{(x - 4)}^{2} {(x + 1)}^{2}} no real zeros \Rightarrow

\Rightarrow no local extremes

y^{″} = \frac{4 (x^{3} - 3 x^{2} + 21 x - 25)}{{(x - 4)}^{3} {(x + 1)}^{3}} = 0 at x \approx 1.33131 \Rightarrow

\Rightarrow turning point

(2)

y = \frac{x^{2} - x - 6}{x^{2} - 3 x + 4} =

= \frac{(x - 3) (x + 2)}{x^{2} - 3 x + 4}; x^{2} - 3 x + 4 = 0 no real zeros \Rightarrow

\Rightarrow {\begin{cases} zeros at x = - 2; x = 3 \\ no vertical asymptote \end{cases}

defined for x \in R

range : transforming y = \frac{x^{2} - x - 6}{x^{2} - 3 x + 4} to

x^{2} - \frac{3 y - 1}{y - 1} x + \frac{2 (2 y + 3)}{y - 1} = 0

D = - \frac{7 y^{2} + 14 y - 25}{4 {(y - 1)}^{2}} ⩾ 0 for - 1 - \frac{4 \sqrt{14}}{7} ⩽ y ⩽ - 1 + \frac{4 \sqrt{14}}{7} \Rightarrow

\Rightarrow {\begin{cases} range = [- 1 - \frac{4 \sqrt{14}}{7}; - 1 + \frac{4 \sqrt{14}}{7}] \\ horizontal asymptote at y = 1 (*) \end{cases}

(*) {doesn}^{'} t mean y = 1 is not within range!!!

x = 5 \Rightarrow y = 1

y^{'} = - \frac{2 (x^{2} - 10 x + 11)}{{(x^{2} - 3 x + 4)}^{2}} = 0 at x = 5 \pm \sqrt{14}

y^{^{″}} = \frac{4 (x^{3} - 15 x^{2} + 33 x - 13)}{{(x^{2} - 3 x + 4)}^{3}}

y^{″} {\begin{cases} > 0 at x = 5 - \sqrt{14} \Rightarrow local minimum \\ < 0 at x = 5 + \sqrt{14} \Rightarrow local maximum \\ = 0 at {\begin{cases} x \approx . 506699 \\ x \approx 2.06421 \\ x \approx 12.4291 \end{cases} \Rightarrow 3 turning points \end{cases}

(3)

y = \frac{x^{2} - x + 6}{x^{2} - 3 x - 4} =

= \frac{x^{2} - x + 6}{(x - 4) (x + 1)} \Rightarrow

\Rightarrow {\begin{cases} no real zeros \\ vertical asymptotes at x = - 1; x = 4 \end{cases}

defined for x \in R ∖ {- 1; 4}

range : transforming y = \frac{x^{2} - x + 6}{x^{2} - 3 x - 4} to

x^{2} - \frac{3 y - 1}{y - 1} x - \frac{2 (2 y + 3)}{y - 1} = 0

D = \frac{25 y^{2} + 2 y - 23}{4 {(y - 1)}^{2}} < 0 for - 1 < y < \frac{23}{25} \Rightarrow

\Rightarrow {\begin{cases} range = R ∖] - 1; \frac{23}{25} [ \\ horizontal asymptote at y = 1 (*) \end{cases}

(*) {doesn}^{'} t mean y = 1 is not within range!!!

x = - 5 \Rightarrow y = 1

y^{'} = - \frac{2 (x^{2} + 10 x - 11)}{{(x - 4)}^{2} {(x + 1)}^{2}} = 0 at x = - 11; x = 1

y^{″} = \frac{4 (x^{3} + 15 x^{2} - 33 x + 53)}{{(x - 4)}^{3} {(x + 1)}^{3}}

y^{″} {\begin{cases} > 0 at x = - 11 \Rightarrow local minimum \\ < 0 at x = 1 \Rightarrow local maximum \\ = 0 at x \approx - 17.1098 \Rightarrow turning point \end{cases}

Commented by TawaTawa last updated on 29/Jan/20

Thanks for your time sir, God bless you .

Commented by jagoll last updated on 29/Jan/20

waw \dots thank you mister . i was

confused, my teacher said that horizontal

asymptotes were not include in the

range .

Commented by jagoll last updated on 29/Jan/20

mr . if given function

y = \frac{ax + b}{cx + d}, is it true that the range

is not the same as the horizontal

asymptote ?

R_{f} : y \neq \frac{a}{c} ?

Commented by john santu last updated on 29/Jan/20

for example

if f (x) = \frac{2 x - 1}{x + 3}

domain x \neq - 3

if y = 2 \Rightarrow 2 = \frac{2 x - 1}{x + 3}

2 x + 6 = 2 x - 1 \Rightarrow 6 = - 1 ? ?

it true for function y = \frac{ax + b}{cx + d}

range y \neq \frac{a}{c}

Commented by jagoll last updated on 29/Jan/20

thanks sir

Commented by MJS last updated on 29/Jan/20

y = \frac{a x + b}{c x + d} \Rightarrow {\begin{cases} zero at x = - \frac{b}{a} \\ vertical asymptote at x = - \frac{d}{c} \end{cases}

defined for x \in R ∖ {- \frac{d}{c}}

y = \frac{a x + b}{c x + d} \Leftrightarrow x = - \frac{d y - b}{c y - a} \Rightarrow range = R ∖ {\frac{a}{c}}

horizontal asymptote at y = \frac{a}{c}

so you are right

Commented by jagoll last updated on 29/Jan/20

sir your notation is

] 14 - 4 \sqrt{6}; 14 + 4 \sqrt{6} [

it same as y ⩽ 14 - 4 \sqrt{6} \lor y ⩾ 14 + 4 \sqrt{6}

?

Commented by MJS last updated on 29/Jan/20

depends \dots

y \in [a; b] is the same as a ⩽ y ⩽ b

\Rightarrow y \in R ∖ [a; b] means y < a \lor y > b

y \in] a; b [is the same as a < y < b

\Rightarrow y \in R ∖] a; b [means y ⩽ a \lor y ⩾ b

Commented by jagoll last updated on 29/Jan/20

thanks you mister

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