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

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

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range of x/(1+x^2)

range

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

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

Interval Notation

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

Solution steps

Rewrite as

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

Multiply both sides by

1 + x^{2}

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

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

Discriminant

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

1 - 4 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 - 14 x - x^{2} d x

- x^{2} + 9

16 x^{3} - x^{4}

f (x) = \frac{x - 1}{x + 4}

(- 4, 3), (4, - 2)