What is the range of (x^2-9x)/(x^2-81) ?

The range of (x^2-9x)/(x^2-81) is f(x) 1

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range of (x^2-9x)/(x^2-81)

range

\frac{x ^{2} - 9 x}{x ^{2} - 81}

f (x) < \frac{1}{2} or \frac{1}{2} < f (x) < 1 or f (x) > 1

Solution steps

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

Domain of

\frac{x ^{2} - 9 x}{x ^{2} - 81} : x < - 9 or - 9 < x < 9 or x > 9

Simplify

\frac{x ^{2} - 9 x}{x ^{2} - 81} : \frac{x}{x + 9}

Extreme Points of

\frac{x}{x + 9} :

None

Find the range for the interval

- \infty < x < - 9 : 1 < f (x) < \infty

Find the range for the interval

- 9 < x < 9 : - \infty < f (x) < \frac{1}{2}

Find the range for the interval

9 < x < \infty : \frac{1}{2} < f (x) < 1

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

f (x) > 1 or f (x) < \frac{1}{2} or \frac{1}{2} < f (x) < 1

f (x) < \frac{1}{2} or \frac{1}{2} < f (x) < 1 or f (x) > 1

(- 1, 7), (3, - 2)

f (x) = \frac{3 x + 9}{x}

sin (2 x - π)

f (x) = ln (\frac{( x + 1 )}{x - 1})

f (x) = 10 t e^{3 - t^{2}}

What is the range of (x^2-9x)/(x^2-81) ?
The range of (x^2-9x)/(x^2-81) is f(x)< 1/2 \lor 1/2 <f(x)<1\lor f(x)>1