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

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

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range of (x^2)/(x+1)

range

\frac{x ^{2}}{x + 1}

f (x) \leq - 4 or f (x) \geq 0

Interval Notation

(- \infty, - 4] \cup [0, \infty)

Solution steps

Rewrite as

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

Multiply both sides by

x + 1

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

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

Discriminant

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

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

Check if the range interval endpoints are included

y = - 4 :

Included

y = 0 :

Included

Therefore the range is

f (x) \leq - 4 or f (x) \geq 0

f (x) = x^{2} - x - 5

\frac{7}{( 2 \cdot 9 + 7 x )}

y = lo g_{5} (x)

- \frac{x + 5}{7}

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

What is the range of (x^2)/(x+1) ?
The range of (x^2)/(x+1) is f(x)<=-4\lor f(x)>= 0