What is the range of (x+6)/(4-sqrt(x^2-9)) ?

The range of (x+6)/(4-sqrt(x^2-9)) is -infinity <f(x)<infinity

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range of (x+6)/(4-sqrt(x^2-9))

range

\frac{x + 6}{4 - x ^{2} - 9}

- \infty < f (x) < \infty

Solution steps

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

Domain of

\frac{x + 6}{4 - x ^{2} - 9} : x < - 5 or - 5 < x \leq - 3 or 3 \leq x < 5 or x > 5

Extreme Points of

\frac{x + 6}{4 - x ^{2} - 9} :

Minimum

(- 3, \frac{3}{4}),

Minimum

(3, \frac{9}{4})

Find the range for the interval

- \infty < x < - 5 : - \infty < f (x) < 1

Find the range for the interval

- 5 < x \leq - 3 : \frac{3}{4} \leq f (x) < \infty

Find the range for the interval

3 \leq x < 5 : \frac{9}{4} \leq f (x) < \infty

Find the range for the interval

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

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

f (x) < 1 or f (x) \geq \frac{3}{4} or f (x) \geq \frac{9}{4} or f (x) < - 1

- \infty < f (x) < \infty

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

f (x) = \frac{x + 5}{x + 1}

x^{2} - 9 x

x^{4} - 6 x^{3}

(x + 5)^{2} + 3

What is the range of (x+6)/(4-sqrt(x^2-9)) ?
The range of (x+6)/(4-sqrt(x^2-9)) is -infinity <f(x)<infinity