What is the range of (4x^2+4x+1)/x ?

The range of (4x^2+4x+1)/x is f(x) = 8

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range of (4x^2+4x+1)/x

range

\frac{4 x ^{2} + 4 x + 1}{x}

f (x) \leq 0 or f (x) \geq 8

Interval Notation

(- \infty, 0] \cup [8, \infty)

Solution steps

Rewrite as

\frac{4 x ^{2} + 4 x + 1}{x} = y

Multiply both sides by

x

4 x^{2} + 4 x + 1 = y x

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

Discriminant

4 x^{2} + 4 x + 1 = y x : y^{2} - 8 y

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

Check if the range interval endpoints are included

y = 0 :

Included

y = 8 :

Included

Therefore the range is

f (x) \leq 0 or f (x) \geq 8

\frac{7}{\frac{x}{x + 7}}

2 x^{2} + 1

f (x) = (x + 8)^{3} + 7

\frac{3}{x + 2}

What is the range of (4x^2+4x+1)/x ?
The range of (4x^2+4x+1)/x is f(x)<= 0\lor f(x)>= 8