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

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

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

range

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

0 < f (x) \leq 2

Interval Notation

(0, 2]

Solution steps

Rewrite as

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

Multiply both sides by

x^{2} + 1

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

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

Discriminant

2 = y (x^{2} + 1) : - 4 y^{2} + 8 y

- 4 y^{2} + 8 y \geq 0 : 0 \leq y \leq 2

Check if the range interval endpoints are included

y = 0 :

Excluded

y = 2 :

Included

Therefore the range is

0 < f (x) \leq 2

f (x) = (4 x - 3)^{\frac{1}{2}}

f (x) = 1 - (x - 3)^{2}

f (x) = \frac{100}{x}

6 x - x^{2} + 7

y = \frac{3}{5}

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