What is the range of f(x)=(x-7)/(x^2+7) ?

The range of f(x)=(x-7)/(x^2+7) is -sqrt(2/7)-1/2 <= f(x)<= sqrt(2/7)-1/2

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range of f(x)=(x-7)/(x^2+7)

range

f (x) = \frac{x - 7}{x ^{2} + 7}

- \frac{2}{7} - \frac{1}{2} \leq f (x) \leq \frac{2}{7} - \frac{1}{2}

Interval Notation

[- \frac{2}{7} - \frac{1}{2}, \frac{2}{7} - \frac{1}{2}]

Solution steps

Rewrite as

\frac{x - 7}{x ^{2} + 7} = y

Multiply both sides by

x^{2} + 7

x - 7 = y (x^{2} + 7)

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

Discriminant

x - 7 = y (x^{2} + 7) : 1 - 28 y^{2} - 28 y

1 - 28 y^{2} - 28 y \geq 0 : - \frac{2}{7} - \frac{1}{2} \leq y \leq \frac{2}{7} - \frac{1}{2}

Check if the range interval endpoints are included

y = - \frac{2}{7} - \frac{1}{2} :

Included

y = \frac{2}{7} - \frac{1}{2} :

Included

Therefore the range is

- \frac{2}{7} - \frac{1}{2} \leq f (x) \leq \frac{2}{7} - \frac{1}{2}

f (x) = \frac{2 x}{- 4 x - 20}

f (x) = 5^{x} - 9

f (x) = \frac{4 x}{x ^{2} + 4}

f (x) = \frac{x ^{2} - 3 x - 4}{x - 2}

y + \frac{1}{2} x = 0

What is the range of f(x)=(x-7)/(x^2+7) ?
The range of f(x)=(x-7)/(x^2+7) is -sqrt(2/7)-1/2 <= f(x)<= sqrt(2/7)-1/2