What is the range of f(x)=(x+4)/(24-sqrt(x^2-49)) ?

The range of f(x)=(x+4)/(24-sqrt(x^2-49)) is f(x) = 11/24

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range of f(x)=(x+4)/(24-sqrt(x^2-49))

range

f (x) = \frac{x + 4}{24 - x ^{2} - 49}

f (x) \leq - \frac{1}{8} or f (x) \geq \frac{11}{24}

Interval Notation

(- \infty, - \frac{1}{8}] \cup [\frac{11}{24}, \infty)

Solution steps

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

Domain of

\frac{x + 4}{24 - x ^{2} - 49} : x < - 25 or - 25 < x \leq - 7 or 7 \leq x < 25 or x > 25

Extreme Points of

\frac{x + 4}{24 - x ^{2} - 49} :

Maximum

(- 7, - \frac{1}{8}),

Minimum

(7, \frac{11}{24})

Find the range for the interval

- \infty < x < - 25 : 1 < f (x) < \infty

Find the range for the interval

- 25 < x \leq - 7 : - \infty < f (x) \leq - \frac{1}{8}

Find the range for the interval

7 \leq x < 25 : \frac{11}{24} \leq f (x) < \infty

Find the range for the interval

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

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

f (x) > 1 or f (x) \leq - \frac{1}{8} or f (x) \geq \frac{11}{24} or f (x) < - 1

f (x) \leq - \frac{1}{8} or f (x) \geq \frac{11}{24}

0.2 x^{2}

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

f (x) = x^{2} - x - 3

\frac{x - 2}{3 x + 7}

f (x) = - 9 x + 2

What is the range of f(x)=(x+4)/(24-sqrt(x^2-49)) ?
The range of f(x)=(x+4)/(24-sqrt(x^2-49)) is f(x)<=-1/8 \lor f(x)>= 11/24