What is the range of f(x)=((3x^2))/(2x+2) ?

The range of f(x)=((3x^2))/(2x+2) is f(x) = 0

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range of f(x)=((3x^2))/(2x+2)

range

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

f (x) \leq - 6 or f (x) \geq 0

Interval Notation

(- \infty, - 6] \cup [0, \infty)

Solution steps

Rewrite as

\frac{3 x ^{2}}{2 x + 2} = y

Multiply both sides by

2 x + 2

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

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

Discriminant

3 x^{2} = y (2 x + 2) : 4 y^{2} + 24 y

4 y^{2} + 24 y \geq 0 : y \leq - 6 or y \geq 0

Check if the range interval endpoints are included

y = - 6 :

Included

y = 0 :

Included

Therefore the range is

f (x) \leq - 6 or f (x) \geq 0

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

(- 1, 8), (- 5, 4)

f (x) = \frac{x ^{2} + 1}{3 ( x - 8 )}

4 x + 8 cos (x)

What is the range of f(x)=((3x^2))/(2x+2) ?
The range of f(x)=((3x^2))/(2x+2) is f(x)<=-6\lor f(x)>= 0