Solve-for-the-domain-and-range-of-f-x-if-f-x-3x-5-x-3-

Question Number 207043 by necx122 last updated on 04/May/24

S o l v e f o r t h e d o m a i n a n d r a n g e o f f (x) i f

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

Answered by Frix last updated on 04/May/24

1 . x + 3 \neq 0

x \neq - 3

2 . \frac{3 x - 5}{x + 3} ⩾ 0

(3 x - 5 < 0 \land x + 3 < 0) \lor (3 x - 5 ⩾ 0 \land x + 3 > 0)

x < - 3 \lor x ⩾ \frac{5}{3} [= domain]

f (x) = y = \sqrt{\frac{3 x - 5}{x + 3}} \Leftrightarrow x = - \frac{3 y^{2} + 5}{y^{2} - 3}

\Rightarrow f (x) ⩾ 0 \land f (x) \neq \sqrt{3} [= range]

Commented by necx122 last updated on 05/May/24

T h a n k y o u s o m u c h .

N o w, f o r g e t t i n g t h e v a l u e s o f t h e

i n e q u a l i t y \frac{3 x - 5}{x + 3} ⩾ 0

w h e n d o y o u d e c i d e t o u s e (3 x - 5 ⩾ 0 \land x + 3 > 0)

a n d (3 x - 5 < 0 \land x + 3 < 0) .

1) C a n t h e s e 2 i n e q u a l i t i e s s a t i s f y t h e

s a m e r a n g e o f v a l u e s ?

2) H o w d o w e c h o o s e w h i c h t o g o w i t h ?

3) w h a t s p e c i f i c r u l e s a r e t h e r e f o r

s o l v i n g r a t i o n a l i n e q u a l i t i e s ?

T h a n k y o u s o m u c h . I a n t i c i p a t e y o u r

r e s p o n s e .

Commented by mr W last updated on 05/May/24

\frac{3 x - 5}{x + 3} ⩾ 0 \Leftrightarrow (3 x - 5) (x + 3) ⩾ 0 \land x + 3 \neq 0

{i t}^{'} s e a s i e r t o t r e a t (3 x - 5) (x + 3) ⩾ 0 .

w e g e t d i r e c t l y x ⩽ - 3 \lor x ⩾ \frac{5}{3} .

s e e d i a g r a m b e l o w .

f o r y ⩾ 0 \Rightarrow x ⩽ x_{1} \lor x ⩾ x_{2}

f o r y ⩽ 0 \Rightarrow x_{1} ⩽ x ⩽ x_{2}

Commented by mr W last updated on 05/May/24

Commented by Frix last updated on 05/May/24

\frac{g (x)}{h (x)} ⩾ 0 means that either both functions

are > 0 or both functions are < 0 .

[\frac{7}{4} > 0 and \frac{- 7}{- 4} is also > 0]

h (x) = 0 is allowed but h (x) = 0 leads to the

undefined \frac{(x)}{0} .

I wrote (g (x) ⩾ 0 \land h (x) > 0) \lor (g (x) < 0 \land h (x) < 0)

where the \lor means “ {or}^{″} so we get 2 regions

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