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

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

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

range

f (x) = \frac{2 x}{x ^{2} + 4}

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

Interval Notation

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

Solution steps

Rewrite as

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

Multiply both sides by

x^{2} + 4

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

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

Discriminant

2 x = y (x^{2} + 4) : 4 - 16 y^{2}

4 - 16 y^{2} \geq 0 : - \frac{1}{2} \leq y \leq \frac{1}{2}

Check if the range interval endpoints are included

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

Included

y = \frac{1}{2} :

Included

Therefore the range is

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

15 x - 6 y = 16

f (x) = - x^{3} + 3 x^{2} - 7

f (x) = x^{2} - 4 x + 7

y = 2 x - 5

e^{- x^{2}}

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