What is the range of (x^2+8x-9)/(x^2+3x-4) ?

The range of (x^2+8x-9)/(x^2+3x-4) is f(x) 2

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range of (x^2+8x-9)/(x^2+3x-4)

range

\frac{x ^{2} + 8 x - 9}{x ^{2} + 3 x - 4}

f (x) < 1 or 1 < f (x) < 2 or f (x) > 2

Interval Notation

(- \infty, 1) \cup (1, 2) \cup (2, \infty)

Solution steps

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

Domain of

\frac{x ^{2} + 8 x - 9}{x ^{2} + 3 x - 4} : x < - 4 or - 4 < x < 1 or x > 1

Simplify

\frac{x ^{2} + 8 x - 9}{x ^{2} + 3 x - 4} : \frac{x + 9}{x + 4}

Extreme Points of

\frac{x + 9}{x + 4} :

None

Find the range for the interval

- \infty < x < - 4 : - \infty < f (x) < 1

Find the range for the interval

- 4 < x < 1 : 2 < f (x) < \infty

Find the range for the interval

1 < x < \infty : 1 < f (x) < 2

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

f (x) < 1 or f (x) > 2 or 1 < f (x) < 2

f (x) < 1 or 1 < f (x) < 2 or f (x) > 2

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

(1, 2), (- 3, 5)

\frac{x}{x + 9}

f (x) = \frac{x + 3}{x - 8}

f (x) = x - 9

What is the range of (x^2+8x-9)/(x^2+3x-4) ?
The range of (x^2+8x-9)/(x^2+3x-4) is f(x)<1\lor 1<f(x)<2\lor f(x)>2