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

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

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range of f(x)=(2x)/(x^2+1)

range

f (x) = \frac{2 x}{x ^{2} + 1}

- 1 \leq f (x) \leq 1

Interval Notation

[- 1, 1]

Solution steps

Rewrite as

\frac{2 x}{x ^{2} + 1} = y

Multiply both sides by

x^{2} + 1

2 x = y (x^{2} + 1)

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

Discriminant

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

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

Check if the range interval endpoints are included

y = - 1 :

Included

y = 1 :

Included

Therefore the range is

- 1 \leq f (x) \leq 1

f (x) = \frac{x + 8}{x ^{2} - 64}

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

sec^{3} (x) d x

y = x + 4

x + 6 y = 42

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