What is the range of 1/(x^2-3x+2) ?

The range of 1/(x^2-3x+2) is f(x) 0

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range of 1/(x^2-3x+2)

range

\frac{1}{x ^{2} - 3 x + 2}

f (x) \leq - 4 or f (x) > 0

Interval Notation

(- \infty, - 4] \cup (0, \infty)

Solution steps

Rewrite as

\frac{1}{x ^{2} - 3 x + 2} = y

Multiply both sides by

x^{2} - 3 x + 2

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

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

Discriminant

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

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

Check if the range interval endpoints are included

y = - 4 :

Included

y = 0 :

Excluded

Therefore the range is

f (x) \leq - 4 or f (x) > 0

f (x) = 2 x + 5

y = (2)^{x} - 3

f (x) = 3 + 2 ln (x)

f (x) = 5 x^{4} + 20 x^{3}

f (x) = - 9 x + 4

What is the range of 1/(x^2-3x+2) ?
The range of 1/(x^2-3x+2) is f(x)<=-4\lor f(x)>0