What is the range of (x-1)/(3x^2-3) ?

The range of (x-1)/(3x^2-3) is f(x) 1/6

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range of (x-1)/(3x^2-3)

range

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

f (x) < 0 or 0 < f (x) < \frac{1}{6} or f (x) > \frac{1}{6}

Interval Notation

(- \infty, 0) \cup (0, \frac{1}{6}) \cup (\frac{1}{6}, \infty)

Solution steps

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

Domain of

\frac{x - 1}{3 x ^{2} - 3} : x < - 1 or - 1 < x < 1 or x > 1

Simplify

\frac{x - 1}{3 x ^{2} - 3} : \frac{1}{3 ( x + 1 )}

Extreme Points of

\frac{1}{3 ( x + 1 )} :

None

Find the range for the interval

- \infty < x < - 1 : - \infty < f (x) < 0

Find the range for the interval

- 1 < x < 1 : \frac{1}{6} < f (x) < \infty

Find the range for the interval

1 < x < \infty : 0 < f (x) < \frac{1}{6}

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

f (x) < 0 or f (x) > \frac{1}{6} or 0 < f (x) < \frac{1}{6}

f (x) < 0 or 0 < f (x) < \frac{1}{6} or f (x) > \frac{1}{6}

x + 5 y = - 10

f (x) = x - 1 + 2

y = - \frac{3}{2} x - 5

f (x) = \frac{1}{x - 3}

f (x) = 12 x - 9

What is the range of (x-1)/(3x^2-3) ?
The range of (x-1)/(3x^2-3) is f(x)<0\lor 0<f(x)< 1/6 \lor f(x)> 1/6