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

The range of x/(2x^2+4) is -1/(4sqrt(2))<= f(x)<= 1/(4sqrt(2))

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

range

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

- \frac{1}{4 2} \leq f (x) \leq \frac{1}{4 2}

Interval Notation

[- \frac{1}{4 2}, \frac{1}{4 2}]

Solution steps

Rewrite as

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

Multiply both sides by

2 x^{2} + 4

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

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

Discriminant

x = y (2 x^{2} + 4) : 1 - 32 y^{2}

1 - 32 y^{2} \geq 0 : - \frac{1}{4 2} \leq y \leq \frac{1}{4 2}

Check if the range interval endpoints are included

y = - \frac{1}{4 2} :

Included

y = \frac{1}{4 2} :

Included

Therefore the range is

- \frac{1}{4 2} \leq f (x) \leq \frac{1}{4 2}

f (x) = 3 x

x^{2} - 6 x + 7

f (x) = - x + 2 + 3

f (x) = \frac{1}{x}

f (x) = sin^{2} (7 x)

What is the range of x/(2x^2+4) ?
The range of x/(2x^2+4) is -1/(4sqrt(2))<= f(x)<= 1/(4sqrt(2))