What is the range of 1/(x^2+2x-8) ?

The range of 1/(x^2+2x-8) is f(x) 0

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range of 1/(x^2+2x-8)

range

\frac{1}{x ^{2} + 2 x - 8}

f (x) \leq - \frac{1}{9} or f (x) > 0

Interval Notation

(- \infty, - \frac{1}{9}] \cup (0, \infty)

Solution steps

Rewrite as

\frac{1}{x ^{2} + 2 x - 8} = y

Multiply both sides by

x^{2} + 2 x - 8

1 = y (x^{2} + 2 x - 8)

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

Discriminant

1 = y (x^{2} + 2 x - 8) : 36 y^{2} + 4 y

36 y^{2} + 4 y \geq 0 : y \leq - \frac{1}{9} or y \geq 0

Check if the range interval endpoints are included

y = - \frac{1}{9} :

Included

y = 0 :

Excluded

Therefore the range is

f (x) \leq - \frac{1}{9} or f (x) > 0

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

f (x) = \frac{7}{x - 9}

f (x) = (x - 4)^{2} - \frac{2}{3} = 6 y - 12

(2.6, 1.3), (1.6, - 5.7)

f (x) = 3 x - 4 y = 12

What is the range of 1/(x^2+2x-8) ?
The range of 1/(x^2+2x-8) is f(x)<=-1/9 \lor f(x)>0