What is the range of (8x)/(x^2+25) ?

The range of (8x)/(x^2+25) is -4/5 <= f(x)<= 4/5

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range of (8x)/(x^2+25)

range

\frac{8 x}{x ^{2} + 25}

- \frac{4}{5} \leq f (x) \leq \frac{4}{5}

Interval Notation

[- \frac{4}{5}, \frac{4}{5}]

Solution steps

Rewrite as

\frac{8 x}{x ^{2} + 25} = y

Multiply both sides by

x^{2} + 25

8 x = y (x^{2} + 25)

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

Discriminant

8 x = y (x^{2} + 25) : 64 - 100 y^{2}

64 - 100 y^{2} \geq 0 : - \frac{4}{5} \leq y \leq \frac{4}{5}

Check if the range interval endpoints are included

y = - \frac{4}{5} :

Included

y = \frac{4}{5} :

Included

Therefore the range is

- \frac{4}{5} \leq f (x) \leq \frac{4}{5}

- 5 x^{3} + 15 x + 4

f (x) = x + 5 - \frac{4 - x}{x}

(- \frac{1}{2}, 0), (0, \frac{1}{4})

f (x) = 12 x - 3 x^{2}

y = \frac{x + 1}{6}

What is the range of (8x)/(x^2+25) ?
The range of (8x)/(x^2+25) is -4/5 <= f(x)<= 4/5