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Question Number 79816 by jagoll last updated on 28/Jan/20

hello mister .

i need help explaining how determine

the range of function

of rational functions

like (i) f (x) = \frac{{ax}^{2} + bx + c}{{px}^{2} + qx + r}

(ii) f (x) = \frac{{ax}^{2} + bx + c}{px + q}

Answered by MJS last updated on 28/Jan/20

y = \frac{{a x}^{2} + b x + c}{{p x}^{2} + q x + r}

leads to

x^{2} + \frac{q y - b}{p y - a} + \frac{r y - c}{p y - a} = 0

D = \frac{(4 p r - q^{2}) y^{2} - 2 (2 a r - b q + 2 c p) y + 4 a c - b^{2}}{4 {(p y - a)}^{2}}

D < 0 \Rightarrow no real solution for x \Rightarrow no y exists

the range then is R ∖ [interval for y where D < 0]

same when p = 0 which is your (ii)

Commented by jagoll last updated on 28/Jan/20

if the function

y = \frac{{2 x}^{2} - 2}{{3 x}^{2} + 3 x - 2} \Rightarrow {3 x}^{2} y + 3 xy - 2 y = {2 x}^{2} - 2

(3 y - 2) . x^{2} + 3 y . x + 2 - 2 y = 0

D = {9 y}^{2} - 4 . (3 y - 2) (2 - 2 y)

D = {9 y}^{2} - 8 (3 y - 2) (1 - y)

D = {9 y}^{2} - 8 (- {3 y}^{2} + 5 y - 2)

D = {33 y}^{2} + 40 y + 16 > 0 for \forall y \in R

R_{f} = {y ∣ y \in R}

that right sir ?

Commented by jagoll last updated on 28/Jan/20

range function not related to

flat asymtot sir ?

Commented by jagoll last updated on 28/Jan/20

for example

y = \frac{{2 x}^{2} + 3 x - 2}{{3 x}^{2} - 5 x + 2}

R_{f} : y \neq \lim_{x \to \infty} \frac{{2 x}^{2} + 3 x - 2}{{3 x}^{2} - 5 x + 2}

R_{f} : y \neq \frac{2}{3} ?

Commented by MJS last updated on 28/Jan/20

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

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

[x^{2} + P x + Q = 0 \Rightarrow D = \frac{P^{2}}{4} - Q]

D = \frac{33 y^{2} - 40 y + 16}{4 {(3 y - 2)}^{2}} > 0 \forall y \in R

\Rightarrow range is R

Commented by MJS last updated on 28/Jan/20

y = \frac{2 x^{2} + 3 x - 2}{3 x^{2} - 5 x + 2}

x^{2} - \frac{5 y + 3}{3 y - 2} x + \frac{2 (y + 1)}{3 y - 2} = 0

D = \frac{y^{2} + 22 y + 25}{4 {(3 y - 2)}^{2}}

D < 0 for - 11 - 4 \sqrt{6} < y < - 11 + 4 \sqrt{6}

\Rightarrow range is R ∖] - 11 - 4 \sqrt{6}; - 11 + 4 \sqrt{6} [

Commented by jagoll last updated on 28/Jan/20

sir how to get in line 2 ? still understand

Commented by jagoll last updated on 28/Jan/20

{3 x}^{2} y + 3 xy - 2 y - {2 x}^{2} + 2 = 0

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

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

that right sir ?

Commented by MJS last updated on 28/Jan/20

y = \frac{2 x^{2} + 3 x - 2}{3 x^{2} - 5 x + 2}

(3 x^{2} - 5 x + 2) y = 2 x^{2} + 3 x - 2

(3 y - 2) x^{2} - (5 y + 3) x + 2 (y + 1) = 0

x^{2} - \frac{5 y + 3}{3 y - 2} x + \frac{2 (y + 1)}{3 y - 2} = 0

{you}^{'} ve made mistakes in line 1

p = - \frac{5 y + 3}{3 y - 2}; q = \frac{2 (y + 1)}{3 y - 2}

D = \frac{p^{2}}{4} - q = \frac{y^{2} + 22 y + 25}{4 {(3 y - 2)}^{2}}

Commented by MJS last updated on 29/Jan/20

look at qu . 79883

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