study-and-give-the-graph-for-the-function-f-x-x-x-1-e-1-x-

Question Number 40106 by maxmathsup by imad last updated on 15/Jul/18

s t u d y a n d g i v e t h e g r a p h f o r t h e f u n c t i o n

f (x) = \frac{x}{x - 1} e^{\frac{1}{x}}

Answered by math khazana by abdo last updated on 28/Jul/18

D_{f} = R - {0, 1} =] - \infty, 0 [\cup] 0, 1 [\cup] 1, + \infty [

{l i m}_{x \to - \infty} f (x) = {l i m}_{x \to - \infty} \frac{x}{x - 1} e^{\frac{1}{x}} = 1

y = 1 i s a s s y m t o t e a t - \infty

{l i m}_{x \to + \infty} f (x) = {l i m}_{x \to + \infty} \frac{x}{x - 1} e^{\frac{1}{x}} = 1 s o

y = 1 i s a s s y m p t o t e a t + \infty

{l i m}_{x \to 1^{-}} f (x) = - \infty a n d {l i m}_{x \to 1^{+}} f (x) = + \infty

{l i m}_{x \to 0^{+}} f (x) = {l i m}_{x \to 0^{+}} \frac{x}{x - 1} e^{\frac{1}{x}} c h a n g = \frac{1}{x} = t

= {l i m}_{t \to + \infty} \frac{1}{t (\frac{1}{t} - 1)} e^{t} = {l i m}_{t \to + \infty} \frac{1}{1 - t} e^{t}

= {l i m}_{t \to + \infty} \frac{e^{t}}{t} \frac{1}{\frac{1}{t} - 1} = - \infty

{l i m}_{x \to 0^{-}} f (x) = {l i m}_{x \to 0^{-}} \frac{x}{x - 1} e^{\frac{1}{x}} (c h . \frac{1}{x} = t)

= {l i m}_{t \to - \infty} \frac{1}{t (\frac{1}{t} - 1)} e^{t} = {l i m}_{t \to - \infty} - \frac{e^{t}}{t} = 0

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

= - \frac{1}{{(x - 1)}^{2}} e^{\frac{1}{x}} - \frac{1}{x^{2}} (\frac{x}{x - 1}) e^{\frac{1}{x}}

= - \frac{1}{{(x - 1)}^{2}} e^{\frac{1}{x}} - \frac{1}{x (x - 1)} e^{\frac{1}{x}}

= - \frac{1}{(x - 1)} e^{\frac{1}{x}} {\frac{1}{x} + 1} = - \frac{x + 1}{x} \frac{1}{(x - 1)} e^{\frac{1}{x}}

= - \frac{x + 1}{x (x - 1)} e^{\frac{1}{x}} \dots . b e c o n t i n u e d \dots