What is the p(x)=((x^3+1))/x+((x^3+8))/2 x^2+3 ?

The p(x)=((x^3+1))/x+((x^3+8))/2 x^2+3 is

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p(x)=((x^3+1))/x+((x^3+8))/2 x^2+3

p (x) = \frac{( x ^{3} + 1 )}{x} + \frac{( x ^{3} + 8 )}{2} x^{2} + 3

Domain : x < 0 or x > 0

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

X Intercepts : (- 0.29195 \dots, 0), (- 2.22597 \dots, 0)

Asymptotes : Vertical x = 0

Interval Notation

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

Range : (- \infty, \infty)

Solution steps

Domain of

\frac{x ^{3} + 1}{x} + \frac{x ^{3} + 8}{2} x^{2} + 3 : x < 0 or x > 0

Range of

\frac{x ^{3} + 1}{x} + \frac{x ^{3} + 8}{2} x^{2} + 3 : - \infty < f (x) < \infty

Axis interception points of

\frac{x ^{3} + 1}{x} + \frac{x ^{3} + 8}{2} x^{2} + 3 :

X Intercepts

: (- 0.29195 \dots, 0), (- 2.22597 \dots, 0)

Asymptotes of

\frac{x ^{3} + 1}{x} + \frac{x ^{3} + 8}{2} x^{2} + 3 :

Vertical

: x = 0

Extreme Points of

\frac{x ^{3} + 1}{x} + \frac{x ^{3} + 8}{2} x^{2} + 3 :

Maximum

Minimum

- 3 x, x < 0

f (x) = \frac{x ^{2} + x - 1}{27}

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

y = 3 x + 2

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

What is the p(x)=((x^3+1))/x+((x^3+8))/2 x^2+3 ?
The p(x)=((x^3+1))/x+((x^3+8))/2 x^2+3 is