What is the range of (2x^2+16x-18)/(x^2+x-6) ?

The range of (2x^2+16x-18)/(x^2+x-6) is -infinity <f(x)<infinity

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range of (2x^2+16x-18)/(x^2+x-6)

range

\frac{2 x ^{2} + 16 x - 18}{x ^{2} + x - 6}

- \infty < f (x) < \infty

Solution steps

Rewrite as

\frac{2 x ^{2} + 16 x - 18}{x ^{2} + x - 6} = y

Multiply both sides by

x^{2} + x - 6

2 x^{2} + 16 x - 18 = y (x^{2} + x - 6)

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

Discriminant

2 x^{2} + 16 x - 18 = y (x^{2} + x - 6) : 25 y^{2} - 152 y + 400

25 y^{2} - 152 y + 400 \geq 0 :

True for all

y \in R

Therefore the range is

- \infty < f (x) < \infty

x^{3} - 3 x

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

f (x) = x^{2} - 3, x \geq 0

\frac{3 x + 4}{2 x - 3}

What is the range of (2x^2+16x-18)/(x^2+x-6) ?
The range of (2x^2+16x-18)/(x^2+x-6) is -infinity <f(x)<infinity