What is the range of (x^3)/(x^2-9) ?

The range of (x^3)/(x^2-9) is -infinity <f(x)<infinity

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range of (x^3)/(x^2-9)

range

\frac{x ^{3}}{x ^{2} - 9}

- \infty < f (x) < \infty

Interval Notation

(- \infty, \infty)

Solution steps

Find the minimum and maximum value in each defined interval and unite the results

Domain of

\frac{x ^{3}}{x ^{2} - 9} : x < - 3 or - 3 < x < 3 or x > 3

Extreme Points of

\frac{x ^{3}}{x ^{2} - 9} :

Maximum

(- 33, - \frac{9 3}{2}),

Saddle

(0, 0),

Minimum

(33, \frac{9 3}{2})

Find the range for the interval

- \infty < x < - 3 : - \infty < f (x) \leq - \frac{9 3}{2}

Find the range for the interval

- 3 < x < 3 : - \infty < f (x) < \infty

Find the range for the interval

3 < x < \infty : \frac{9 3}{2} \leq f (x) < \infty

Combine the ranges of all domain intervals to obtain the function range

f (x) \leq - \frac{9 3}{2} or - \infty < f (x) < \infty or f (x) \geq \frac{9 3}{2}

- \infty < f (x) < \infty

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

\frac{3}{x}

\frac{2}{x ^{2} - 16}

3 - 2 x

sec (x)

What is the range of (x^3)/(x^2-9) ?
The range of (x^3)/(x^2-9) is -infinity <f(x)<infinity