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

The range of f(x)=(x-3)/(x^2-1) is f(x) = sqrt(2)+3/2

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

range

f (x) = \frac{x - 3}{x ^{2} - 1}

f (x) \leq - 2 + \frac{3}{2} or f (x) \geq 2 + \frac{3}{2}

Interval Notation

(- \infty, - 2 + \frac{3}{2}] \cup [2 + \frac{3}{2}, \infty)

Solution steps

Rewrite as

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

Multiply both sides by

x^{2} - 1

x - 3 = y (x^{2} - 1)

The range is the set of y for which the discriminant is greater or equal to zero

Discriminant

x - 3 = y (x^{2} - 1) : 1 + 4 y^{2} - 12 y

1 + 4 y^{2} - 12 y \geq 0 : y \leq - 2 + \frac{3}{2} or y \geq 2 + \frac{3}{2}

Check if the range interval endpoints are included

y = - 2 + \frac{3}{2} :

Included

y = 2 + \frac{3}{2} :

Included

Therefore the range is

f (x) \leq - 2 + \frac{3}{2} or f (x) \geq 2 + \frac{3}{2}

f (x) = -, 4 \leq x \leq 5

(- 6, - 10), (- 2, - 8)

f (x) = \frac{5}{x ^{2} - 5 x}

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

f (x) = y + 1 = 3 (x - 4)

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