range of (sqrt(x+2))/(6x^2+x-2)

Solution

range

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

f (x) \leq - \frac{( 911 + 23 1969 ) 23 + 1969}{23520} or f (x) \geq 0

Interval Notation

(- \infty, - \frac{( 911 + 23 1969 ) 23 + 1969}{23520}] \cup [0, \infty)

Solution steps

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

Domain of

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

Extreme Points of

\frac{x + 2}{6 x ^{2} + x - 2} :

Minimum

(- 2, 0),

Maximum

(- \frac{49 - 1969}{36}, - \frac{( 911 + 23 1969 ) 23 + 1969}{23520})

Find the range for the interval

- 2 \leq x < - \frac{2}{3} : 0 \leq f (x) < \infty

Find the range for the interval

- \frac{2}{3} < x < \frac{1}{2} : - \infty < f (x) \leq - \frac{( 911 + 23 1969 ) 23 + 1969}{23520}

Find the range for the interval

\frac{1}{2} < x < \infty : 0 < f (x) < \infty

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

f (x) \geq 0 or f (x) \leq - \frac{( 911 + 23 1969 ) 23 + 1969}{23520} or f (x) > 0

f (x) \leq - \frac{( 911 + 23 1969 ) 23 + 1969}{23520} or f (x) \geq 0

ln (2 + ln (x))

2^{3 x - 5}

x^{2} + y^{2} + 4 y - 60 = 0

lo g_{\frac{1}{2}} (- x) + 2

x^{3} - x^{2} - 8 x