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Popular >

f(x)=(x(1+ln|x|))/((1-x))

Solution

f (x) = \frac{x ( 1 + ln ∣ x ∣ )}{( 1 - x )}

Solution

Domain : x < 0 or 0 < x < 1 or x > 1

Range : - \infty < f (x) < \infty

X Intercepts : (- \frac{1}{e}, 0), (\frac{1}{e}, 0)

Asymptotes : Vertical x = 1

Extreme Points : Maximum (- W_{0} (\frac{1}{e ^{2}}), - \frac{W _{0} ( \frac{1}{e ^{2}} ) ( 1 + ln ( W _{0} ( \frac{1}{e ^{2}} ) ) )}{1 + W _{0} ( \frac{1}{e ^{2}} )}), Minimum (- W_{0} (- \frac{1}{e ^{2}}), - \frac{W _{0} ( - \frac{1}{e ^{2}} ) ( 1 + ln ( - W _{0} ( - \frac{1}{e ^{2}} ) ) )}{1 + W _{0} ( - \frac{1}{e ^{2}} )}), Maximum (- W_{- 1} (- \frac{1}{e ^{2}}), \frac{W _{- 1} ( - \frac{1}{e ^{2}} ) ( 1 + ln ( - W _{- 1} ( - \frac{1}{e ^{2}} ) ) )}{- W _{- 1} ( - \frac{1}{e ^{2}} ) - 1})

+1

Interval Notation

Domain : (- \infty, 0) \cup (0, 1) \cup (1, \infty)

Range : (- \infty, \infty)

Solution steps

Domain of

\frac{x ( 1 + ln ∣ x ∣ )}{1 - x} : x < 0 or 0 < x < 1 or x > 1

Range of

\frac{x ( 1 + ln ∣ x ∣ )}{1 - x} : - \infty < f (x) < \infty

Axis interception points of

\frac{x ( 1 + ln ∣ x ∣ )}{1 - x} :

X Intercepts

: (- \frac{1}{e}, 0), (\frac{1}{e}, 0)

Asymptotes of

\frac{x ( 1 + ln ∣ x ∣ )}{1 - x} :

Vertical

: x = 1

Extreme Points of

\frac{x ( 1 + ln ∣ x ∣ )}{1 - x} :

Maximum

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

Minimum

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

Maximum

(- W_{- 1} (- \frac{1}{e ^{2}}), \frac{W _{- 1} ( - \frac{1}{e ^{2}} ) ( 1 + ln ( - W _{- 1} ( - \frac{1}{e ^{2}} ) ) )}{- W _{- 1} ( - \frac{1}{e ^{2}} ) - 1})

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Frequently Asked Questions (FAQ)

What is the f(x)=(x(1+ln|x|))/((1-x)) ?
The f(x)=(x(1+ln|x|))/((1-x)) is