What is the range of f(x)=([(4x^2+1)])/(2x) ?

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

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range of f(x)=([(4x^2+1)])/(2x)

range

f (x) = \frac{[ ( 4 x ^{2} + 1 ) ]}{2 x}

f (x) \leq - 2 or f (x) \geq 2

Interval Notation

(- \infty, - 2] \cup [2, \infty)

Solution steps

Rewrite as

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

Multiply both sides by

x

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

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

Discriminant

\frac{4 x ^{2} + 1}{2} = y x : 4 y^{2} - 16

4 y^{2} - 16 \geq 0 : y \leq - 2 or y \geq 2

Check if the range interval endpoints are included

y = - 2 :

Included

y = 2 :

Included

Therefore the range is

f (x) \leq - 2 or f (x) \geq 2

\frac{x + 4}{x ^{2} + 7 x + 12}

f (x) = ln (x) + 2

f (x) = x^{4}

f (x) = \frac{5 - x}{x ^{2} - 4}

f (x) = \frac{1}{x} - 3

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