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

The range of (x+5)/(x^2+x-6) is f(x) = (2sqrt(14)-9)/(25)

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

range

\frac{x + 5}{x ^{2} + x - 6}

f (x) \leq \frac{- 2 14 - 9}{25} or f (x) \geq \frac{2 14 - 9}{25}

Solution steps

Rewrite as

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

Multiply both sides by

x^{2} + x - 6

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

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

Discriminant

x + 5 = y (x^{2} + x - 6) : 25 y^{2} + 18 y + 1

25 y^{2} + 18 y + 1 \geq 0 : y \leq \frac{- 2 14 - 9}{25} or y \geq \frac{2 14 - 9}{25}

Check if the range interval endpoints are included

y = \frac{- 2 14 - 9}{25} :

Included

y = \frac{2 14 - 9}{25} :

Included

Therefore the range is

f (x) \leq \frac{- 2 14 - 9}{25} or f (x) \geq \frac{2 14 - 9}{25}

y = lo g_{\frac{1}{5}} (x)

f (x) = \frac{4 x}{9} - \frac{28}{9}

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

f (x) = - 3 x + 1

What is the range of (x+5)/(x^2+x-6) ?
The range of (x+5)/(x^2+x-6) is f(x)<= (-2sqrt(14)-9)/(25)\lor f(x)>= (2sqrt(14)-9)/(25)