What is the range of f(x)= 1/(x^2+19) ?

The range of f(x)= 1/(x^2+19) is 0<f(x)<= 1/19

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range of f(x)= 1/(x^2+19)

range

f (x) = \frac{1}{x ^{2} + 19}

0 < f (x) \leq \frac{1}{19}

Interval Notation

(0, \frac{1}{19}]

Solution steps

Rewrite as

\frac{1}{x ^{2} + 19} = y

Multiply both sides by

x^{2} + 19

1 = y (x^{2} + 19)

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

Discriminant

1 = y (x^{2} + 19) : - 76 y^{2} + 4 y

- 76 y^{2} + 4 y \geq 0 : 0 \leq y \leq \frac{1}{19}

Check if the range interval endpoints are included

y = 0 :

Excluded

y = \frac{1}{19} :

Included

Therefore the range is

0 < f (x) \leq \frac{1}{19}

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

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

\frac{5}{x - 7}

(- 1, 8), (3, - 6)

f (x) = - x^{5} - 1

What is the range of f(x)= 1/(x^2+19) ?
The range of f(x)= 1/(x^2+19) is 0<f(x)<= 1/19