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

The range of (sqrt(y^2-1))/y is -1<f(y)<1

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range of (sqrt(y^2-1))/y

range

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

- 1 < f (y) < 1

Interval Notation

(- 1, 1)

Solution steps

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

Domain of

\frac{y ^{2} - 1}{y} : y \leq - 1 or y \geq 1

Extreme Points of

\frac{y ^{2} - 1}{y} :

Maximum

(- 1, 0),

Minimum

(1, 0)

Find the range for the interval

- \infty < y \leq - 1 : - 1 < f (y) \leq 0

Find the range for the interval

1 \leq y < \infty : 0 \leq f (y) < 1

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

- 1 < f (y) \leq 0 or 0 \leq f (y) < 1

- 1 < f (y) < 1

(1, 4), (4, 8)

f (x) = - \frac{3}{4} x + \frac{15}{4}

f (x) = 7 x^{2} + 2

f (x) = \frac{4}{x - 5} + 3

\frac{5}{x - 4}

What is the range of (sqrt(y^2-1))/y ?
The range of (sqrt(y^2-1))/y is -1<f(y)<1