What is the range of f(x)= 5/(x^2-9x-22) ?

The range of f(x)= 5/(x^2-9x-22) is f(x) 0

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range of f(x)= 5/(x^2-9x-22)

range

f (x) = \frac{5}{x ^{2} - 9 x - 22}

f (x) \leq - \frac{20}{169} or f (x) > 0

Interval Notation

Solution steps

Rewrite as

\frac{5}{x ^{2} - 9 x - 22} = y

Multiply both sides by

x^{2} - 9 x - 22

5 = y (x^{2} - 9 x - 22)

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

Discriminant

5 = y (x^{2} - 9 x - 22) : 169 y^{2} + 20 y

169 y^{2} + 20 y \geq 0 : y \leq - \frac{20}{169} or y \geq 0

Check if the range interval endpoints are included

y = - \frac{20}{169} :

Included

y = 0 :

Excluded

Therefore the range is

f (x) \leq - \frac{20}{169} or f (x) > 0

f (x) = 14 x^{2}

(10, 242), (15, 364)

f (x) = y

f (x) = sin (\frac{8 π x}{5})

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

What is the range of f(x)= 5/(x^2-9x-22) ?
The range of f(x)= 5/(x^2-9x-22) is f(x)<=-20/169 \lor f(x)>0