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

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

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range of 1/(x^2+4)

range

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

0 < f (x) \leq \frac{1}{4}

Interval Notation

(0, \frac{1}{4}]

Solution steps

Rewrite as

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

Multiply both sides by

x^{2} + 4

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

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

Discriminant

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

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

Check if the range interval endpoints are included

y = 0 :

Excluded

y = \frac{1}{4} :

Included

Therefore the range is

0 < f (x) \leq \frac{1}{4}

f (x) = \frac{x}{5}

f (x) = arccos (2 x)

5 x + 8 y = 16

f (x) = - x^{2} - 6 x + 1

y = - 4 x + 1