What is the f(x)=(6x^5+4x-2x^2-7)/(5x^2) ?

The f(x)=(6x^5+4x-2x^2-7)/(5x^2) is

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f(x)=(6x^5+4x-2x^2-7)/(5x^2)

f (x) = \frac{6 x ^{5} + 4 x - 2 x ^{2} - 7}{5 x ^{2}}

Domain : x < 0 or x > 0

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

X Intercepts : (0.96429 \dots, 0)

Asymptotes : Vertical x = 0

Extreme Points : Maximum (- 1, - \frac{19}{5})

Interval Notation

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

Range : (- \infty, \infty)

Solution steps

Domain of

\frac{6 x ^{5} + 4 x - 2 x ^{2} - 7}{5 x ^{2}} : x < 0 or x > 0

Range of

\frac{6 x ^{5} + 4 x - 2 x ^{2} - 7}{5 x ^{2}} : - \infty < f (x) < \infty

Axis interception points of

\frac{6 x ^{5} + 4 x - 2 x ^{2} - 7}{5 x ^{2}} :

X Intercepts

: (0.96429 \dots, 0)

Asymptotes of

\frac{6 x ^{5} + 4 x - 2 x ^{2} - 7}{5 x ^{2}} :

Vertical

: x = 0

Extreme Points of

\frac{6 x ^{5} + 4 x - 2 x ^{2} - 7}{5 x ^{2}} :

Maximum

(- 1, - \frac{19}{5})

y = arccos (sin (x))

f (t) = t + 1

y = - 9 x + 2

f (θ) = \frac{1 + cos ( θ ) + sin ( θ )}{1 + cos ( θ ) - sin ( θ )}

f (x) = \frac{1}{x ^{2} + 5}