What is the f(y)=yln(1+e^y) ?

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f(y)=yln(1+e^y)

f (y) = y ln (1 + e^{y})

Domain : - \infty < y < \infty

X Intercepts : (0, 0), Y Intercepts : (0, 0)

Asymptotes : Horizontal y = 0

Inflection Points : (- W_{0} (\frac{2}{e ^{2}}) - 2, ln (1 + e^{- W_{0} (\frac{2}{e ^{2}}) - 2}) (- W_{0} (\frac{2}{e ^{2}}) - 2))

Interval Notation

Domain : (- \infty, \infty)

Solution steps

Domain of

y ln (1 + e^{y}) : - \infty < y < \infty

Axis interception points of

y ln (1 + e^{y}) :

X Intercepts

: (0, 0),

Y Intercepts

: (0, 0)

Asymptotes of

y ln (1 + e^{y}) :

Horizontal

: y = 0

Inflection Points of

y ln (1 + e^{y}) : (- W_{0} (\frac{2}{e ^{2}}) - 2, ln (1 + e^{- W_{0} (\frac{2}{e ^{2}}) - 2}) (- W_{0} (\frac{2}{e ^{2}}) - 2))

\int \frac{cos ( x )}{sin ^{3} ( x ) - cos ^{3} ( x )} d x

\frac{d}{d x} (sin (2 x) + 2 x cos (2 x))

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

\int \frac{x ^{2}}{( 16 - x ^{3} ) ^{2}} d x

x \to - 4 + lim (\frac{x + 3}{x + 4})