What is the range of f(x)=((x-1))/(x(x^2-9)) ?

The range of f(x)=((x-1))/(x(x^2-9)) is -infinity <f(x)<infinity

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range of f(x)=((x-1))/(x(x^2-9))

range

f (x) = \frac{( x - 1 )}{x ( x ^{2} - 9 )}

- \infty < f (x) < \infty

Solution steps

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

Domain of

\frac{x - 1}{x ( x ^{2} - 9 )} : x < - 3 or - 3 < x < 0 or 0 < x < 3 or x > 3

Extreme Points of

\frac{x - 1}{x ( x ^{2} - 9 )} :

Maximum

(- 1.27372 \dots, - 0.24196 \dots)

Find the range for the interval

- \infty < x < - 3 : 0 < f (x) < \infty

Find the range for the interval

- 3 < x < 0 : - \infty < f (x) \leq - 0.24196 \dots

Find the range for the interval

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

Find the range for the interval

3 < x < \infty : 0 < f (x) < \infty

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

f (x) > 0 or f (x) \leq - 0.24196 \dots or - \infty < f (x) < \infty or f (x) > 0

- \infty < f (x) < \infty

f (x) = - 4 x^{2} + 12

5 t + 6

- \frac{1}{2} x^{2} + 2 x + 3

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

f (x) = \frac{9 x + 4}{x - 1}

What is the range of f(x)=((x-1))/(x(x^2-9)) ?
The range of f(x)=((x-1))/(x(x^2-9)) is -infinity <f(x)<infinity