What is the range of x/(2x^2+9x+4) ?

The range of x/(2x^2+9x+4) is f(x) = (4sqrt(2)+9)/(49)

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range of x/(2x^2+9x+4)

range

\frac{x}{2 x ^{2} + 9 x + 4}

f (x) \leq \frac{- 4 2 + 9}{49} or f (x) \geq \frac{4 2 + 9}{49}

Interval Notation

(- \infty, \frac{- 4 2 + 9}{49}] \cup [\frac{4 2 + 9}{49}, \infty)

Solution steps

Rewrite as

\frac{x}{2 x ^{2} + 9 x + 4} = y

Multiply both sides by

2 x^{2} + 9 x + 4

x = y (2 x^{2} + 9 x + 4)

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

Discriminant

x = y (2 x^{2} + 9 x + 4) : 49 y^{2} - 18 y + 1

49 y^{2} - 18 y + 1 \geq 0 : y \leq \frac{- 4 2 + 9}{49} or y \geq \frac{4 2 + 9}{49}

Check if the range interval endpoints are included

y = \frac{- 4 2 + 9}{49} :

Included

y = \frac{4 2 + 9}{49} :

Included

Therefore the range is

f (x) \leq \frac{- 4 2 + 9}{49} or f (x) \geq \frac{4 2 + 9}{49}

- 2 x + y = 4

f (x) = x^{7} + 7 x^{3} + 1

y = - 4 x - 6, (9, 4)

f (x) = arccos (x)

What is the range of x/(2x^2+9x+4) ?
The range of x/(2x^2+9x+4) is f(x)<= (-4sqrt(2)+9)/(49)\lor f(x)>= (4sqrt(2)+9)/(49)