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

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

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

range

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

f (x) < 0 or 0 < f (x) < \frac{1}{2} or f (x) > \frac{1}{2}

Solution steps

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

Domain of

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

Simplify

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

Extreme Points of

\frac{1}{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}{2} < f (x) < \infty

Find the range for the interval

1 < x < \infty : 0 < f (x) < \frac{1}{2}

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

f (x) < 0 or f (x) > \frac{1}{2} or 0 < f (x) < \frac{1}{2}

f (x) < 0 or 0 < f (x) < \frac{1}{2} or f (x) > \frac{1}{2}

y = x^{2} - 4 x + 3

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

- x^{4} + 2 x^{3} + 4 x^{2}

y = \frac{3}{4} x + 3

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