What is the range of g(x)= 1/(x^2+8) ?

The range of g(x)= 1/(x^2+8) is 0<f(x)<= 1/8

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range of g(x)= 1/(x^2+8)

range

g (x) = \frac{1}{x ^{2} + 8}

0 < f (x) \leq \frac{1}{8}

Interval Notation

(0, \frac{1}{8}]

Solution steps

Rewrite as

\frac{1}{x ^{2} + 8} = y

Multiply both sides by

x^{2} + 8

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

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

Discriminant

1 = y (x^{2} + 8) : - 32 y^{2} + 4 y

- 32 y^{2} + 4 y \geq 0 : 0 \leq y \leq \frac{1}{8}

Check if the range interval endpoints are included

y = 0 :

Excluded

y = \frac{1}{8} :

Included

Therefore the range is

0 < f (x) \leq \frac{1}{8}

234

f (x) = x^{3} - 27 x, - 4 \leq x \leq 4

y = 4.5

tan^{2} (x)

\frac{x ^{3}}{3} - 2 x^{2} - 5 x

What is the range of g(x)= 1/(x^2+8) ?
The range of g(x)= 1/(x^2+8) is 0<f(x)<= 1/8