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Question Number 36442 by abdo mathsup 649 cc last updated on 02/Jun/18

l e t f (x) = \sqrt{2 + x^{2}} - x

1) c a l c u l a t e {l i m}_{x \to + \infty} f (x) a n d {l i m}_{x \to - \infty} f (x)

2) c a l c u l a t e {l i m}_{x \to + \infty} \frac{f (x)}{x} a n d {l i m}_{x \to - \infty} \frac{f (x)}{x}

3) c a l c u l a t e f^{^{'}} (x) a n d d e t e r m i n e i t s s i g n

4) g i v e t h e v a r i a t i o n o f

5) g i v e t h e e q u a t i o n o f a s s y m p t o t e a t p o i n t

A (1, f (1))

6) f i n d f^{- 1} (x) a n d c a l c u l a t e {(f^{- 1})}^{^{'}} (x)

7) c a l c u l a t e \int_{0}^{4} f (x) d x .

Commented by abdo.msup.com last updated on 05/Jun/18

1) {l i m}_{x \to + \infty} f (x) = {l i m}_{x \to + \infty} \frac{2}{\sqrt{2 + x^{2}} + x}

= 0 a n d {l i m}_{x \to - \infty} f (x) = + \infty

2) {l i m}_{x \to + \infty} \frac{f (x)}{x} = l i m \frac{\sqrt{2 + x^{2}} - x}{x}

= {l i m}_{x \to + \infty} \sqrt{\frac{2}{x^{2}} + 1} - 1 = 0

{l i m}_{x \to - \infty} \frac{f (x)}{x} = {l i m}_{x \to - \infty} \frac{\sqrt{2 + x^{2}} - x}{x}

= {l i m}_{x \to - \infty} - \sqrt{\frac{2}{x^{2}} + 1} - 1 = - 2

3) w e h a v e f (x) = \sqrt{2 + x^{2}} - x \Rightarrow

f^{^{'}} (x) = \frac{2 x}{2 \sqrt{2 + x^{2}}} - 1 = \frac{x}{\sqrt{2 + x^{2}}} - 1

= \frac{x - \sqrt{2 + x^{2}}}{\sqrt{2 + x^{2}}} i f x ⩽ 0 f^{^{'}} (x) ⩽ 0

i f x ⩾ 0 w e h a v e x^{2} - {(\sqrt{2 + x^{2}})}^{2} ⩽ 0 \Rightarrow

f^{^{'}} (x) ⩽ 0

4) f i s d e c r e a s i n g o n R .

Commented by abdo.msup.com last updated on 05/Jun/18

5) w e h a v e f (x) = \sqrt{2 + x^{2}} - x \Rightarrow

f (1) = \sqrt{3} - 1 a l s o f^{^{'}} (x) = \frac{x}{\sqrt{2 + x^{2}}} - 1 \Rightarrow

f^{^{'}} (1) = \frac{1}{\sqrt{3}} - 1 s o t h e e q u s t i o n o f

a s s y m p t o t e i s

y = f^{^{'}} (1) (x - 1) + f (1) \Rightarrow

y = (\frac{1}{\sqrt{3}} - 1) (x - 1) + \sqrt{3} - 1 .

Commented by abdo.msup.com last updated on 05/Jun/18

6) f i s d e c r e a s i n g s o f] - \infty, + \infty [

=] f (+ \infty), f (- \infty) [=] 0, + \infty [f i s

a b i j e c t i o n f r o m R t o] 0, + \infty [

l e t f (x) = y \Leftrightarrow \sqrt{2 + x^{2}} - x = y \Rightarrow

\sqrt{2 + x^{2}} = x + y (s o x + y > 0) \Rightarrow

2 + x^{2} = x^{2} + 2 x y + y^{2} \Rightarrow

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

\Rightarrow f^{- 1} (x) = \frac{2 - x^{2}}{2 x} a n d x > 0

Commented by abdo.msup.com last updated on 05/Jun/18

w e h a v e f^{- 1} (x) = \frac{1}{x} - \frac{x}{2} \Rightarrow

{(f^{- 1} (x))}^{^{'}} = - \frac{1}{x^{2}} - \frac{1}{2} .

Commented by abdo.msup.com last updated on 05/Jun/18

7) l e t A = \int_{0}^{4} f (x) d x

A = \int_{0}^{4} (\sqrt{2 + x^{2}} - x) d x

= \int_{0}^{4} \sqrt{2 + x^{2}} d x - {[\frac{x^{2}}{2}]}_{0}^{4}

= \int_{0}^{4} \sqrt{2 + x^{2}} d x - 8 b u t c h a n g . x = \sqrt{2} s h (t)

g i v e

\int_{0}^{4} \sqrt{2 + x^{2}} d x = \sqrt{2} \int_{0}^{a r g s h (\frac{4}{\sqrt{2}})} c h (t) \sqrt{2} c h t d t

= 2 \int_{0}^{a r g s h (\frac{4}{\sqrt{2}})} {c h}^{2} (t) d t = 2 \int_{0}^{a r g s h (\frac{4}{\sqrt{2}})} \frac{1 + c h (2 t)}{2} d t

= a r g s h (\frac{4}{\sqrt{2}}) + \frac{1}{2} {[s h (2 t)]}_{0}^{a r g s h (\frac{4}{\sqrt{2}})}

= l n (\frac{4}{\sqrt{2}} + 3) + \frac{1}{2} s h (2 l n {\frac{4}{\sqrt{2}} + 3)})

= l n (\frac{4}{\sqrt{2}} + 3) + \frac{1}{4} {{(\frac{4}{\sqrt{2}} + 3)}^{2} + {(\frac{4}{\sqrt{2}} + 3)}^{- 1}} .

Commented by abdo.msup.com last updated on 05/Jun/18

I = l n (4 + 3 \sqrt{2}) - \frac{1}{2} l n (2)

+ \frac{1}{4} {{(\frac{4}{\sqrt{2}} + 3)}^{2} - {(\frac{4}{\sqrt{2}} + 3)}^{- 2}} - 8 .