What is the range of f(x)=((x^2-25))/(x^2-4x-5) ?

The range of f(x)=((x^2-25))/(x^2-4x-5) is f(x) 5/3

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range of f(x)=((x^2-25))/(x^2-4x-5)

range

f (x) = \frac{( x ^{2} - 25 )}{x ^{2} - 4 x - 5}

f (x) < 1 or 1 < f (x) < \frac{5}{3} or f (x) > \frac{5}{3}

Interval Notation

(- \infty, 1) \cup (1, \frac{5}{3}) \cup (\frac{5}{3}, \infty)

Solution steps

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

Domain of

\frac{x ^{2} - 25}{x ^{2} - 4 x - 5} : x < - 1 or - 1 < x < 5 or x > 5

Simplify

\frac{x ^{2} - 25}{x ^{2} - 4 x - 5} : \frac{x + 5}{x + 1}

Extreme Points of

\frac{x + 5}{x + 1} :

None

Find the range for the interval

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

Find the range for the interval

- 1 < x < 5 : \frac{5}{3} < f (x) < \infty

Find the range for the interval

5 < x < \infty : 1 < f (x) < \frac{5}{3}

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

f (x) < 1 or f (x) > \frac{5}{3} or 1 < f (x) < \frac{5}{3}

f (x) < 1 or 1 < f (x) < \frac{5}{3} or f (x) > \frac{5}{3}

f (x) = cos (x)

1200 + 500 sin (0.3 t)

f (x) = x^{2} - 1

\frac{1}{x ^{2} - 4}

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

What is the range of f(x)=((x^2-25))/(x^2-4x-5) ?
The range of f(x)=((x^2-25))/(x^2-4x-5) is f(x)<1\lor 1<f(x)< 5/3 \lor f(x)> 5/3