What is the range of 1/(x^2-7x+10) ?

The range of 1/(x^2-7x+10) is f(x) 0

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range of 1/(x^2-7x+10)

range

\frac{1}{x ^{2} - 7 x + 10}

f (x) \leq - \frac{4}{9} or f (x) > 0

Interval Notation

(- \infty, - \frac{4}{9}] \cup (0, \infty)

Solution steps

Rewrite as

\frac{1}{x ^{2} - 7 x + 10} = y

Multiply both sides by

x^{2} - 7 x + 10

1 = y (x^{2} - 7 x + 10)

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

Discriminant

1 = y (x^{2} - 7 x + 10) : 9 y^{2} + 4 y

9 y^{2} + 4 y \geq 0 : y \leq - \frac{4}{9} or y \geq 0

Check if the range interval endpoints are included

y = - \frac{4}{9} :

Included

y = 0 :

Excluded

Therefore the range is

f (x) \leq - \frac{4}{9} or f (x) > 0

f (x) = \frac{1}{x - 4}

x (3 - x)

f (x) = \frac{e ^{- 2 x}}{x - 7}

9 - 3^{x}

What is the range of 1/(x^2-7x+10) ?
The range of 1/(x^2-7x+10) is f(x)<=-4/9 \lor f(x)>0