NEW FUNCTION

Function Expression :

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

Domain

\[\left]-\infty, -1\right[ \cup \left]-1, 0\right[ \cup \left]0, \infty\right[ \]

Limits

\[\lim_{x \rightarrow-\infty}f(x) = 1 \]
\[\lim_{x \overset{<}{\rightarrow-1} }f(x) = +\infty \]
\[\lim_{x \overset{>}{\rightarrow-1} }f(x) = -\infty \]
\[\lim_{x \overset{<}{\rightarrow0} }f(x) = -\infty \]
\[\lim_{x \overset{>}{\rightarrow0} }f(x) = 0 \]
\[\lim_{x \rightarrow+\infty}f(x) = 1 \]
\[ \]

Derivate

\[f^{\,\prime}(x)=- \frac{\left(x - 1\right) e^{- \frac{1}{x}}}{\left(x + 1\right)^{2}} + \frac{e^{- \frac{1}{x}}}{x + 1} + \frac{\left(x - 1\right) e^{- \frac{1}{x}}}{x^{2} \left(x + 1\right)} \]
\[f^{\,\prime}(x)=\frac{\left(3 x^{2} - 1\right) e^{- \frac{1}{x}}}{x^{2} \left(x^{2} + 2 x + 1\right)} \]
\[ \]

Integral

\[F(x) = \int \frac{\left(x - 1\right) e^{- \frac{1}{x}}}{x + 1}\, dx \]

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