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Question Number 35986 by abdo mathsup 649 cc last updated on 26/May/18

l e t f (x) = \sqrt{1 + n x^{2}} - n x + 3 w i t h n i n t e g r

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

2) c a l c u l a t e f^{^{'}} (x)

3) 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 o f f a t

p o i n t A (1, f (1)) .

4) 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} .

Commented by prof Abdo imad last updated on 31/Aug/18

1) w e h a v e {l i m}_{x \to - \infty} (- n x + 3) = + \infty a n d

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

f o r x > 0 \sqrt{1 + {n x}^{2}} = \sqrt{{n x}^{2} (1 + \frac{1}{{n x}^{2}}})

= x \sqrt{n} \sqrt{1 + \frac{1}{{n x}^{2}}} \sim x \sqrt{n} {1 + \frac{1}{2 {n x}^{2}}} = x \sqrt{n} + \frac{\sqrt{n}}{2 n x}

= x \sqrt{n} + \frac{1}{2 x \sqrt{n}} (x \to + \infty) \Rightarrow

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

f (x) \sim - \sqrt{n} (\sqrt{n} - 1) x \Rightarrow {l i m}_{x \to + \infty} f (x) = - \infty

Commented by prof Abdo imad last updated on 31/Aug/18

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

= \frac{n x}{\sqrt{1 + {n x}^{2}}} - n = n {\frac{x}{\sqrt{1 + {n x}^{2}}} - 1} .

Commented by prof Abdo imad last updated on 31/Aug/18

3) w e h a v e f (1) = \sqrt{1 + n} - n + 3 a n d

f^{^{'}} (x) = \frac{n x}{\sqrt{1 + {n x}^{2}}} - n \Rightarrow f^{^{'}} (1) = \frac{n}{\sqrt{1 + n}} - n s o 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 i s

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

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

Commented by prof Abdo imad last updated on 31/Aug/18

4) w e h a v e f o r x > 0

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

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

f o r x < 0 \frac{f (x)}{x} = - \sqrt{\frac{1 + {n x}^{2}}{x^{2}}} - n + \frac{3}{x}

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

Commented by maxmathsup by imad last updated on 31/Aug/18

3) f o r g i v e e q u a t i o n o f t a n g e n t e \dots .