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

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

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

range

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

0 < f (x) \leq 1

Interval Notation

Solution steps

Rewrite as

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

Multiply both sides by

1 + x^{2}

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

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

Discriminant

1 = y (1 + x^{2}) : - 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}{( x - 2 ) ( x + 3 )}

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

f (x) = 5 x - 3

\frac{x ^{2} + 5}{2}

h (x) = \frac{5}{7} x^{5} - 3

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