What is the f(x)=(x^4}{12}+\frac{x^3)/6-x^2+x ?

The f(x)=(x^4}{12}+\frac{x^3)/6-x^2+x is

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f(x)=(x^4}{12}+\frac{x^3)/6-x^2+x

f (x) = \frac{x ^{4}}{12} + \frac{x ^{3}}{6} - x^{2} + x

Domain : - \infty < x < \infty

X Intercepts : (0, 0), (- 4.92870 \dots, 0), Y Intercepts : (0, 0)

Asymptotes : None

Extreme Points : Minimum (- 3.47503 \dots, - 10.39269 \dots), Maximum (0.65302 \dots, 0.28815 \dots), Minimum (1.32201 \dots, 0.21392 \dots)

Interval Notation

Domain : (- \infty, \infty)

Solution steps

Domain of

\frac{x ^{4}}{12} + \frac{x ^{3}}{6} - x^{2} + x : - \infty < x < \infty

Axis interception points of

\frac{x ^{4}}{12} + \frac{x ^{3}}{6} - x^{2} + x :

X Intercepts

: (0, 0), (- 4.92870 \dots, 0),

Y Intercepts

: (0, 0)

Asymptotes of

\frac{x ^{4}}{12} + \frac{x ^{3}}{6} - x^{2} + x :

None

Extreme Points of

\frac{x ^{4}}{12} + \frac{x ^{3}}{6} - x^{2} + x :

Minimum

(- 3.47503 \dots, - 10.39269 \dots),

Maximum

(0.65302 \dots, 0.28815 \dots),

Minimum

(1.32201 \dots, 0.21392 \dots)

f (x) = \frac{x ^{6}}{2}

y = - 2 (x + 3) (x - 2)

f (x) = \frac{x + 6}{x ^{2} - 9 x + 18}

f (x) = x^{3} - x^{2} + 3

f (x) = [cos (x^{3} + 1)] 3

What is the f(x)=(x^4}{12}+\frac{x^3)/6-x^2+x ?
The f(x)=(x^4}{12}+\frac{x^3)/6-x^2+x is