range of (x^3-x)/(x^2-6x+5)

Solution

range

\frac{x ^{3} - x}{x ^{2} - 6 x + 5}

f (x) \leq - \frac{60 - 11 30}{30} or f (x) \geq \frac{60 + 11 30}{30}

Interval Notation

(- \infty, - \frac{60 - 11 30}{30}] \cup [\frac{60 + 11 30}{30}, \infty)

Solution steps

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

Domain of

\frac{x ^{3} - x}{x ^{2} - 6 x + 5} : x < 1 or 1 < x < 5 or x > 5

Simplify

\frac{x ^{3} - x}{x ^{2} - 6 x + 5} : \frac{x ( x + 1 )}{x - 5}

Extreme Points of

\frac{x ( x + 1 )}{x - 5} :

Maximum

(5 - 30, - \frac{60 - 11 30}{30}),

Minimum

(5 + 30, \frac{60 + 11 30}{30})

Find the range for the interval

- \infty < x < 1 : - \infty < f (x) \leq - \frac{60 - 11 30}{30}

Find the range for the interval

1 < x < 5 : - \infty < f (x) < - \frac{1}{2}

Find the range for the interval

5 < x < \infty : \frac{60 + 11 30}{30} \leq f (x) < \infty

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

f (x) \leq - \frac{60 - 11 30}{30} or f (x) < - \frac{1}{2} or f (x) \geq \frac{60 + 11 30}{30}

f (x) \leq - \frac{60 - 11 30}{30} or f (x) \geq \frac{60 + 11 30}{30}

x - 3 = 0

y = 11 x^{2} - 7 x - 24

f (x) = \frac{2}{x}

f (x) = 1. 1^{x}

x^{4} - 50 x^{2} + 5