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Question Number 36316 by ajfour last updated on 31/May/18

F i n d d o m a i n a n d r a n g e o f t h e

f u n c t i o n f (x) = \frac{x^{2} - 6 x + 8}{x - 5} .

A l s o d r a w t h e g r a p h .

Answered by tanmay.chaudhury50@gmail.com last updated on 31/May/18

y = \frac{x^{2} - 6 x + 8}{x - 5}

x^{2} - 6 x + 8 - x y + 5 y = 0

x^{2} + x (- 6 - y) + 8 + 5 y = 0

d i s c r i m i n a n t

D = B^{2} - 4 A C

= {(6 + y)}^{2} - 4 (8 + 5 y)

= 36 + 12 y + y^{2} - 32 - 20 y

= y^{2} - 8 y + 4

= y^{2} - 2 . y . 4 + {(4)}^{2} + 4 - 16

= {(y - 4)}^{2} - {(2 \sqrt{3})}^{2}

= {(y - 4)}^{2} - {(2 \sqrt{3})}^{2}

= (y - 4 + 2 \sqrt{3}) (y - 4 - 2 \sqrt{3})

c r i t i c a l v a l u e o f y a r e (4 - 2 \sqrt{3}), (4 + 2 \sqrt{3})

D > 0 w h e n y > 4 + 2 \sqrt{3}

D > 0 w h e n y < (4 - 2 \sqrt{3}

s o y c a n n o t h a v e v a l u e b e t w e e n (4 - 2 \sqrt{3})

a n d (4 + 2 \sqrt{3})

v a l u e o f y

(- \infty, 4 - 2 \sqrt{3}] \cup [4 + 2 \sqrt{3}, + \infty)

a t x = 5 y i s u n d e f i n e d s o x c a n n o t b e 5

p l s c h e c k a n d s u g g e s t

d o m a i n o f x = R - {5}

Commented by tanmay.chaudhury50@gmail.com last updated on 31/May/18

l a t e r c o r r e c t e d a n d p u t i n r e d c o l o u r e d f o n t

Commented by tanmay.chaudhury50@gmail.com last updated on 31/May/18

i h a v e d o n e a m i s t a k e i n p l a c e o f 8 + 5 y i p u t

8 - 5 y

Answered by ajfour last updated on 31/May/18

Commented by ajfour last updated on 31/May/18

\frac{d f (x)}{d x} = \frac{(2 x - 6) (x - 5) - (x^{2} - 6 x + 8)}{{(x - 5)}^{2}}

\Rightarrow f^{'} (x) = 0 f o r

x^{2} - 10 x + 22 = 0

\Rightarrow x = \frac{10 \pm \sqrt{100 - 88}}{2}

= 5 \pm \sqrt{3} (p o i n t o f l o c a l

m i n i m a a n d m a x i m a)

f (x) = \frac{(x - 2) (x - 4)}{(x - 5)}

f (5 - \sqrt{3}) = \frac{(3 - \sqrt{3}) (1 - \sqrt{3})}{- \sqrt{3}}

= \frac{4 \sqrt{3} - 6}{\sqrt{3}} = 4 - 2 \sqrt{3}

f (5 + \sqrt{3}) = \frac{(3 + \sqrt{3}) (1 + \sqrt{3})}{\sqrt{3}}

= 4 + 2 \sqrt{3} .

D o m a i n : x \in R - {5}

R a n g e :

f (x) \in (- \infty, 4 - 2 \sqrt{3}] \cup [4 + 2 \sqrt{3}, \infty) .

Answered by ajfour last updated on 31/May/18

x y - 5 y = x^{2} - 6 x + 8

x^{2} - (y + 6) x + 5 y + 8 = 0

f o r r e a l x

{(y + 6)}^{2} ⩾ 4 (5 y + 8)

y^{2} - 8 y + 4 ⩾ 0

\Rightarrow r o o t s a r e y = 4 \pm 2 \sqrt{3}

\Rightarrow y \in (- \infty, 4 - 2 \sqrt{3}] \cup [4 + 2 \sqrt{3}, \infty) .

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