What is the range of f(x)=(3x^2)/(2x^2-32) ?

The range of f(x)=(3x^2)/(2x^2-32) is f(x) 3/2

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range of f(x)=(3x^2)/(2x^2-32)

range

f (x) = \frac{3 x ^{2}}{2 x ^{2} - 32}

f (x) \leq 0 or f (x) > \frac{3}{2}

Interval Notation

Solution steps

Rewrite as

\frac{3 x ^{2}}{2 x ^{2} - 32} = y

Multiply both sides by

2 x^{2} - 32

3 x^{2} = y (2 x^{2} - 32)

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} - 32) : 256 y^{2} - 384 y

256 y^{2} - 384 y \geq 0 : y \leq 0 or y \geq \frac{3}{2}

Check if the range interval endpoints are included

y = 0 :

Included

y = \frac{3}{2} :

Excluded

Therefore the range is

f (x) \leq 0 or f (x) > \frac{3}{2}

f (x) = \frac{( 4 x - 8 )}{( x + 6 )}

\frac{d y}{y}

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

y = 4 x + 5

f (x) = 5

What is the range of f(x)=(3x^2)/(2x^2-32) ?
The range of f(x)=(3x^2)/(2x^2-32) is f(x)<= 0\lor f(x)> 3/2