What is the range of f(x)=((6x^2-5))/((2x^2+6)) ?

The range of f(x)=((6x^2-5))/((2x^2+6)) is -5/6 <= f(x)<3

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range of f(x)=((6x^2-5))/((2x^2+6))

range

f (x) = \frac{( 6 x ^{2} - 5 )}{( 2 x ^{2} + 6 )}

- \frac{5}{6} \leq f (x) < 3

Solution steps

Rewrite as

\frac{6 x ^{2} - 5}{2 x ^{2} + 6} = y

Multiply both sides by

2 x^{2} + 6

6 x^{2} - 5 = y (2 x^{2} + 6)

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

Discriminant

6 x^{2} - 5 = y (2 x^{2} + 6) : - 48 y^{2} + 104 y + 120

- 48 y^{2} + 104 y + 120 \geq 0 : - \frac{5}{6} \leq y \leq 3

Check if the range interval endpoints are included

y = - \frac{5}{6} :

Included

y = 3 :

Excluded

Therefore the range is

- \frac{5}{6} \leq f (x) < 3

\frac{( 1 )}{( - 10 ( \frac{( 1 )}{( - 5 x - 6 )} ) + 3 )}

3 x^{2}

(- 4) (6)

\frac{1 - 4 t}{3 + t}

f (x) = \frac{x + 4}{( x - 3 ) ( x + 4 )}

What is the range of f(x)=((6x^2-5))/((2x^2+6)) ?
The range of f(x)=((6x^2-5))/((2x^2+6)) is -5/6 <= f(x)<3