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Question Number 38721 by maxmathsup by imad last updated on 28/Jun/18

l e t f (x) = \sqrt{1 + 2 x^{2}} - x \sqrt{2} + 3

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) g i v e t h e a s s y m t o t e t o g r a p h C_{f}

4) g i v e t h e a s s y m t o t e t o C_{f} a t p o i n t A (0, f (0))

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

6) c a l c u l a t e \int_{0}^{1} f (x) d x .

Commented by prof Abdo imad last updated on 02/Jul/18

1) w e h a v e \sqrt{1 + 2 x^{2}} = x \sqrt{2} \sqrt{1 + \frac{1}{2 x^{2}}} \sim

x \sqrt{2} {1 + \frac{1}{4 x^{2}}} (x \to + \infty) \Rightarrow f (x) \sim \frac{\sqrt{2}}{4 x} + 3 \Rightarrow

{l i m}_{x \to + \infty} f (x) = 3 a l s o w e c a n u s e t h i s m e t h o d

{l i m}_{x \to + \infty} f (x) = l i m \frac{(\sqrt{1 + 2 x^{2}} - (x \sqrt{2} - 3)) (\sqrt{1 + 2 x^{2}} + x \sqrt{2} - 3)}{\sqrt{1 + 2 x^{2}} + x \sqrt{2} - 3}

= {l i m}_{x \to + \infty} \frac{1 + 2 x^{2} - (2 x^{2} - 6 x \sqrt{2} + 9)}{\sqrt{1 + 2 x^{2}} + x \sqrt{2} - 3}

= {l i m}_{x \to + \infty} \frac{6 x \sqrt{2} - 8}{\sqrt{1 + 2 x^{2}} + x \sqrt{2} - 3}

= {l i m}_{x \to + \infty} \frac{6 x \sqrt{2} - 8}{x \sqrt{2} {\sqrt{1 + \frac{1}{2 x^{2}}} + 1 - \frac{3}{x \sqrt{2}}}} = 3

{l i m}_{x \to - \infty} f (x) = {l i m}_{x \to - \infty} \sqrt{1 + 2 x^{2}} - x \sqrt{2} + 3 = + \infty

2) w e h a v e {l i m}_{x \to + *} f (x) = 3 \Rightarrow {l i m}_{x \to + \infty} \frac{f (x)}{x} = 0

\sqrt{1 + 2 x^{2}} = - x \sqrt{2} \sqrt{1 + \frac{1}{2 x^{2}}} f o r x < 0

\sim - x \sqrt{2} (1 + \frac{1}{4 x^{2}}) (x \to - \infty) \Rightarrow

f (x) \sim - 2 \sqrt{2} x - \frac{\sqrt{2}}{4 x} + 3 \Rightarrow \frac{f (x)}{x} \sim - 2 \sqrt{2} - \frac{\sqrt{2}}{4 x^{2}} + \frac{3}{x}

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

Commented by prof Abdo imad last updated on 02/Jul/18

{l i m}_{x \to - \infty} \frac{f (x)}{x} = - 2 \sqrt{2} b u t w e h a v e

f (x) \sim - 2 \sqrt{2} x + 3 - \frac{\sqrt{2}}{4 x} (x \to - \infty) \Rightarrow

{l i m}_{x \to - \infty} f (x) - (- 2 \sqrt{2} x + 3) = 0 s o

y = - 2 \sqrt{2} + 3 i s e q u a t i o n o f a s s h m p t o t e t o C_{f}

a t - \infty

Commented by prof Abdo imad last updated on 02/Jul/18

4) f i r s t c h a n g e a s s y m p t o t e b y e q u a t i o n o f

t a n g e n t

w e h a v e f^{^{'}} (x) = \frac{4 x}{2 \sqrt{1 + 2 x^{2}}} - \sqrt{2} = \frac{2 x}{\sqrt{1 + 2 x^{2}}} - \sqrt{2}

\Rightarrow f^{^{'}} (0) = - \sqrt{2}

f (0) = 4 \Rightarrow t h e e q u a t i o n o f a s s y m p t o t e i s

y = f^{^{'}} (0) x + f (0) = - \sqrt{2} x + 4

Commented by prof Abdo imad last updated on 02/Jul/18

5) f (x) = y \Leftrightarrow x = f^{- 1} (y) \Rightarrow \sqrt{1 + 2 x^{2}} - x \sqrt{2} + 3 = y

\Rightarrow \sqrt{1 + 2 x^{2}} = (x \sqrt{2} + y - 3) \Rightarrow

1 + 2 x^{2} = {(x \sqrt{2} + y - 3)}^{2} \Rightarrow

1 + 2 x^{2} = 2 x^{2} + 2 x \sqrt{2} (y - 3) + y^{2} - 6 y + 9 \Rightarrow

2 \sqrt{2} (y - 3) x + y^{2} - 6 y + 8 = 0 \Rightarrow

2 \sqrt{2} (y - 3) x = - y^{2} + 6 y - 8 \Rightarrow

x = \frac{- y^{2} + 6 y - 8}{2 \sqrt{2} (y - 3)} = f^{- 1} (y) \Rightarrow

f^{- 1} (x) = \frac{x^{2} - 6 x + 8}{2 \sqrt{2} (3 - x)} .

{(f^{- 1} (x))}^{^{'}} = \frac{1}{2 \sqrt{2}} {\frac{(2 x - 6) (3 - x) + (x^{2} - 6 x + 8)}{{(3 - x)}^{2}}}

= \frac{1}{2 \sqrt{2}} {\frac{6 x - 2 x^{2} - 18 + 6 x + x^{2} - 6 x + 8}{{(3 - x)}^{2}}}

= \frac{1}{2 \sqrt{2}} {\frac{- x^{2} + 6 x - 10}{{(3 - x)}^{2}}}