x-4-x-1-gt-4x-5-

Question Number 180569 by SAMIRA last updated on 13/Nov/22

\sqrt{x + 4} - \sqrt{x - 1} > \sqrt{4 x + 5}

Commented by Frix last updated on 14/Nov/22

false .

the inequality is defined for x ⩾ 1

but for x ⩾ 0 : \sqrt{x + 4} < \sqrt{4 x + 5}

\Rightarrow

for x ⩾ 1 \sqrt{x + 4} < \sqrt{4 x + 5} \land \sqrt{x - 1} ⩾ 0

\Rightarrow \sqrt{x + 4} - \sqrt{x - 1} < \sqrt{4 x + 5}

Answered by Acem last updated on 14/Nov/22

f (x) = \sqrt{4 x + 5}; x \in [- \frac{5}{4}, + \infty [

c_{1} (- 1.25, 0), c_{2} (0, \sqrt{5})

g (x) = \sqrt{x + 4} - \sqrt{x - 1}; x \in [1, + \infty [

c_{3} (1, \sqrt{5}),

\lim_{x \to + \infty} g (x) = \lim_{x \to + \infty} \frac{(\sqrt{x + 4} - \sqrt{x - 1}) (\sqrt{x + 4} + \sqrt{x - 1})}{\sqrt{x + 4} + \sqrt{x - 1}}

= \lim_{x \to + \infty} \frac{5}{\sqrt{x + 4} + \sqrt{x - 1}} = 0

\Rightarrow f o r x \in [1, + \infty [: g (x) \in [\sqrt{5}, 0 [

& f (x) \in [3, + \infty [

\Rightarrow \sqrt{4 x + 5} i s a l w a y s b i g g e r t h a n g (x) \forall x \in [1, + \infty [

T h e n t h e s t a t e m e n t

\sqrt{x + 4} - \sqrt{x - 1} > \sqrt{4 x + 5} i s w r o n g o r x \in \emptyset

Commented by Acem last updated on 13/Nov/22

Facebook Tweet Pin