What is the range of f(x)=(-3x-5)/(2x^2-4x) ?

The range of f(x)=(-3x-5)/(2x^2-4x) is f(x) = (sqrt(55))/4+2

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range of f(x)=(-3x-5)/(2x^2-4x)

range

f (x) = \frac{- 3 x - 5}{2 x ^{2} - 4 x}

f (x) \leq - \frac{55}{4} + 2 or f (x) \geq \frac{55}{4} + 2

Interval Notation

(- \infty, - \frac{55}{4} + 2] \cup [\frac{55}{4} + 2, \infty)

Solution steps

Rewrite as

\frac{- 3 x - 5}{2 x ^{2} - 4 x} = y

Multiply both sides by

2 x^{2} - 4 x

- 3 x - 5 = y (2 x^{2} - 4 x)

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

Discriminant

- 3 x - 5 = y (2 x^{2} - 4 x) : 16 y^{2} - 64 y + 9

16 y^{2} - 64 y + 9 \geq 0 : y \leq - \frac{55}{4} + 2 or y \geq \frac{55}{4} + 2

Check if the range interval endpoints are included

y = - \frac{55}{4} + 2 :

Included

y = \frac{55}{4} + 2 :

Included

Therefore the range is

f (x) \leq - \frac{55}{4} + 2 or f (x) \geq \frac{55}{4} + 2

f (x) = \frac{x ^{2} + 3 x + 2}{- 3 x - 12}

f (x) = \frac{6}{x + 5} + x

x^{2} + y^{2} = 25

f (x) = \frac{1}{3} x^{3} - x^{2} - 4

7 cos (10 x - 6 π) + 8

What is the range of f(x)=(-3x-5)/(2x^2-4x) ?
The range of f(x)=(-3x-5)/(2x^2-4x) is f(x)<=-(sqrt(55))/4+2\lor f(x)>= (sqrt(55))/4+2