What is the range of f(x)= 4/(x^2+1) ?

The range of f(x)= 4/(x^2+1) is 0<f(x)<= 4

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range of f(x)= 4/(x^2+1)

range

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

0 < f (x) \leq 4

Interval Notation

Solution steps

Rewrite as

\frac{4}{x ^{2} + 1} = y

Multiply both sides by

x^{2} + 1

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

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

Discriminant

4 = y (x^{2} + 1) : - 4 y^{2} + 16 y

- 4 y^{2} + 16 y \geq 0 : 0 \leq y \leq 4

Check if the range interval endpoints are included

y = 0 :

Excluded

y = 4 :

Included

Therefore the range is

0 < f (x) \leq 4

f (x) = ln ∣ 2 x - 3 ∣

49 - x^{2}

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

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

x^{2} - 2 x - 15

What is the range of f(x)= 4/(x^2+1) ?
The range of f(x)= 4/(x^2+1) is 0<f(x)<= 4