find-the-range-f-x-2-6-x-2-

Question Number 73649 by aliesam last updated on 14/Nov/19

f i n d t h e r a n g e

f (x) = \frac{2}{6 - \sqrt{x + 2}}

Commented by mathmax by abdo last updated on 14/Nov/19

x \in D_{f} \Leftrightarrow x ⩾ - 2 a n d \sqrt{x + 2} \neq 6 \Leftrightarrow x ⩾ - 2 a n d x + 2 \neq 36 \Leftrightarrow x ⩾ - 2

a n d x \neq 34 \Rightarrow D_{f} = [- 2, 34 [\cup] 34, + \infty [

f^{^{'}} (x) = 2 (- \frac{- \frac{1}{2 \sqrt{x + 2}}}{{(6 - \sqrt{x + 2})}^{2}}) = \frac{1}{\sqrt{x + 2} {(6 - \sqrt{x + 2})}^{2}} > 0 \Rightarrow f i s i n c r e a z i n g

o n D_{f} w e h a v e f (- 2) = \frac{1}{3} a n d {l i m}_{x \to 34^{-}} f (x) = + \infty

{l i m}_{x \to 34^{+}} f (x) = - \infty \Rightarrow f ([- 2, 34 [) = [\frac{1}{3}, + \infty [a n d

f (] 34, + \infty [) =] - \infty, 0 [v a r i a t i o n o f f

x - 2 34 + \infty

f^{^{'}} + ∣∣ +

f \frac{1}{3} i n c + \infty ∣∣ - \infty i n c 0

Commented by malwaan last updated on 14/Nov/19

x ⩾ - 2 a n d x \neq 34

\Rightarrow - 2 ⩽ x < 34 \lor x > 34

- 2 ⩽ x < 34 \Rightarrow 0 ⩽ x + 2 < 36

\Rightarrow 0 ⩽ \sqrt{x + 2} < 6 \Rightarrow 0 ⩾ - \sqrt{x + 2} > - 6

\Rightarrow 6 ⩾ 6 - \sqrt{x + 2} > 0

\Rightarrow \frac{1}{3} ⩽ \frac{2}{6 - \sqrt{x + 2}} < \infty

\Rightarrow \frac{1}{3} ⩽ y < \infty \dots . (1)

x > 34 \Rightarrow 34 < x < \infty

\Rightarrow 36 < x + 2 < \infty

\Rightarrow 6 < \sqrt{x + 2} < \infty

\Rightarrow - 6 > - \sqrt{x + 2} > - \infty

\Rightarrow 0 > 6 - \sqrt{x + 2} > - \infty

\Rightarrow - \infty < \frac{2}{6 - \sqrt{x + 2}} < 0

\Rightarrow - \infty < y < 0 \dots . (2)

f r o m (1); (2)

\Rightarrow t h e r a n g e =

[1 ╱ 3; \infty [\cup] - \infty; 0 [

Answered by MJS last updated on 14/Nov/19

defined for x ⩾ - 2 \land x \neq 34

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

\lim_{x \to 34^{-}} f (x) = + \infty

\lim_{x \to 34^{+}} f (x) = - \infty

\lim_{x \to + \infty} f (x) = 0

\Rightarrow - \infty < y ⩽ 0 \lor \frac{1}{3} ⩽ y < + \infty

Commented by aliesam last updated on 14/Nov/19

g o d b l e s s y o u s i r

Commented by aliesam last updated on 14/Nov/19

b u t i t h i n k t h a t

- \infty < y < 0

Commented by MJS last updated on 14/Nov/19

yes you are right, {it}^{'} s a typo

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