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

The range of (x^2-2)/(x+2) is f(x) = 2sqrt(2)-4

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range of (x^2-2)/(x+2)

range

\frac{x ^{2} - 2}{x + 2}

f (x) \leq - 22 - 4 or f (x) \geq 22 - 4

Interval Notation

(- \infty, - 22 - 4] \cup [22 - 4, \infty)

Solution steps

Rewrite as

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

Multiply both sides by

x + 2

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

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

Discriminant

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

y^{2} + 8 + 8 y \geq 0 : y \leq - 22 - 4 or y \geq 22 - 4

Check if the range interval endpoints are included

y = - 22 - 4 :

Included

y = 22 - 4 :

Included

Therefore the range is

f (x) \leq - 22 - 4 or f (x) \geq 22 - 4

3 y = 5 x + 7

y = 3 tan (4 x + 6) - 3

\frac{75 p}{85 - p}

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

x^{3} - 12 x + 1

What is the range of (x^2-2)/(x+2) ?
The range of (x^2-2)/(x+2) is f(x)<=-2sqrt(2)-4\lor f(x)>= 2sqrt(2)-4