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

The range of f(x)= 1/(x^2+1) is 0<f(x)<= 1

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

range

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

0 < f (x) \leq 1

Interval Notation

(0, 1]

Solution steps

Rewrite as

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

Multiply both sides by

x^{2} + 1

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

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

Discriminant

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

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

Check if the range interval endpoints are included

y = 0 :

Excluded

y = 1 :

Included

Therefore the range is

0 < f (x) \leq 1

f (x) = \frac{( x ^{2} - 16 )}{( x - 2 )}

\frac{5 x ^{3} - 11}{9}

(2) (4)

f (x) = 2 x^{3} + 12 x^{2} - 30 x

- \frac{1}{3} cos (\frac{1}{3} x)

What is the range of f(x)= 1/(x^2+1) ?
The range of f(x)= 1/(x^2+1) is 0<f(x)<= 1