What is the range of f(x)=(x^2-5)/(x^2-9) ?

The range of f(x)=(x^2-5)/(x^2-9) is f(x) 1

Popular >

range of f(x)=(x^2-5)/(x^2-9)

range

f (x) = \frac{x ^{2} - 5}{x ^{2} - 9}

f (x) \leq \frac{5}{9} or f (x) > 1

Interval Notation

(- \infty, \frac{5}{9}] \cup (1, \infty)

Solution steps

Rewrite as

\frac{x ^{2} - 5}{x ^{2} - 9} = y

Multiply both sides by

x^{2} - 9

x^{2} - 5 = y (x^{2} - 9)

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

Discriminant

x^{2} - 5 = y (x^{2} - 9) : 36 y^{2} - 56 y + 20

36 y^{2} - 56 y + 20 \geq 0 : y \leq \frac{5}{9} or y \geq 1

Check if the range interval endpoints are included

y = \frac{5}{9} :

Included

y = 1 :

Excluded

Therefore the range is

f (x) \leq \frac{5}{9} or f (x) > 1

\frac{1}{4} x^{4} - 4 x^{3} + 24 x^{2}

f (x) = \frac{x - 3}{x ^{2} + 15 x + 56}

lo g_{9} (x)

f (x) = (x + 1 \circ x^{2} - 1)

- 3 x^{4} + 24 x^{3} - 48 x^{2}

What is the range of f(x)=(x^2-5)/(x^2-9) ?
The range of f(x)=(x^2-5)/(x^2-9) is f(x)<= 5/9 \lor f(x)>1