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

The range of 1/(-x^2+2x+3) is f(x) = 1/4

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

range

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

f (x) < 0 or f (x) \geq \frac{1}{4}

Interval Notation

(- \infty, 0) \cup [\frac{1}{4}, \infty)

Solution steps

Rewrite as

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

Multiply both sides by

- x^{2} + 2 x + 3

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

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

Discriminant

1 = y (- x^{2} + 2 x + 3) : 16 y^{2} - 4 y

16 y^{2} - 4 y \geq 0 : y \leq 0 or y \geq \frac{1}{4}

Check if the range interval endpoints are included

y = 0 :

Excluded

y = \frac{1}{4} :

Included

Therefore the range is

f (x) < 0 or f (x) \geq \frac{1}{4}

f (x) = \frac{5 x + 4}{x ^{2} + 3 x + 2}

f (x) = 2 x - 3 y = 12

\frac{x + 8}{x ^{2} - 64}

g (x) = - \frac{2}{x} - 1

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