What is the range of (27x)/(x^2+9) ?

The range of (27x)/(x^2+9) is -9/2 <= f(x)<= 9/2

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range of (27x)/(x^2+9)

range

\frac{27 x}{x ^{2} + 9}

- \frac{9}{2} \leq f (x) \leq \frac{9}{2}

Interval Notation

[- \frac{9}{2}, \frac{9}{2}]

Solution steps

Rewrite as

\frac{27 x}{x ^{2} + 9} = y

Multiply both sides by

x^{2} + 9

27 x = y (x^{2} + 9)

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

Discriminant

27 x = y (x^{2} + 9) : 729 - 36 y^{2}

729 - 36 y^{2} \geq 0 : - \frac{9}{2} \leq y \leq \frac{9}{2}

Check if the range interval endpoints are included

y = - \frac{9}{2} :

Included

y = \frac{9}{2} :

Included

Therefore the range is

- \frac{9}{2} \leq f (x) \leq \frac{9}{2}

5^{x} + 3

(- 3, - 2), (2, 3)

(x^{2} - 7)^{3}

f (x) = \frac{4}{3 + x}

What is the range of (27x)/(x^2+9) ?
The range of (27x)/(x^2+9) is -9/2 <= f(x)<= 9/2