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

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

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

range

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

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

Interval Notation

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

Solution steps

Rewrite as

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

Multiply both sides by

x^{2} + 1

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

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

Discriminant

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

f (x) = \frac{x - 4}{x ^{2} + 10 x + 21}

f (x) = \frac{( 2 x + 3 )}{( x - 4 )}

- 5 + ln (x)

2 x^{3} + x^{2} - 3 x + 1

y = x + 7

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