What is the range of f(x)= 4/(x^2-5x+6) ?

The range of f(x)= 4/(x^2-5x+6) is f(x) 0

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range of f(x)= 4/(x^2-5x+6)

range

f (x) = \frac{4}{x ^{2} - 5 x + 6}

f (x) \leq - 16 or f (x) > 0

Interval Notation

(- \infty, - 16] \cup (0, \infty)

Solution steps

Rewrite as

\frac{4}{x ^{2} - 5 x + 6} = y

Multiply both sides by

x^{2} - 5 x + 6

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

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

Discriminant

4 = y (x^{2} - 5 x + 6) : y^{2} + 16 y

y^{2} + 16 y \geq 0 : y \leq - 16 or y \geq 0

Check if the range interval endpoints are included

y = - 16 :

Included

y = 0 :

Excluded

Therefore the range is

f (x) \leq - 16 or f (x) > 0

y = x - 2

f (x) = (x + 4)^{\frac{1}{3}}

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

f (x) = - 5 x + 3

f (x) = x^{2} - 14 x + 49

What is the range of f(x)= 4/(x^2-5x+6) ?
The range of f(x)= 4/(x^2-5x+6) is f(x)<=-16\lor f(x)>0