What is the range of (12-x-x^2)/(x-3) ?

The range of (12-x-x^2)/(x-3) is f(x) -7

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range of (12-x-x^2)/(x-3)

range

\frac{12 - x - x ^{2}}{x - 3}

f (x) < - 7 or f (x) > - 7

Interval Notation

(- \infty, - 7) \cup (- 7, \infty)

Solution steps

Find the minimum and maximum value in each defined interval and unite the results

Domain of

\frac{12 - x - x ^{2}}{x - 3} : x < 3 or x > 3

Simplify

\frac{12 - x - x ^{2}}{x - 3} : - x - 4

Extreme Points of

- x - 4 :

None

Find the range for the interval

- \infty < x < 3 : - 7 < f (x) < \infty

Find the range for the interval

3 < x < \infty : - \infty < f (x) < - 7

Combine the ranges of all domain intervals to obtain the function range

f (x) > - 7 or f (x) < - 7

f (x) < - 7 or f (x) > - 7

f (x) = x^{2} + y - 16 = 0

\frac{x ^{2} - 6 x - 72}{x ^{2} - 18 x + 72}

f (x) = 4 sin (50 x)

f (x) = - x^{2} + 9 x - 8 - 4

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

What is the range of (12-x-x^2)/(x-3) ?
The range of (12-x-x^2)/(x-3) is f(x)<-7\lor f(x)>-7