What is the range of (2x-5)/(sqrt(x^2-3x-28)) ?

The range of (2x-5)/(sqrt(x^2-3x-28)) is f(x) = (6sqrt(13))/(11)

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range of (2x-5)/(sqrt(x^2-3x-28))

range

\frac{2 x - 5}{x ^{2} - 3 x - 28}

f (x) < - 2 or f (x) \geq \frac{6 13}{11}

Interval Notation

(- \infty, - 2) \cup [\frac{6 13}{11}, \infty)

Solution steps

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

Domain of

\frac{2 x - 5}{x ^{2} - 3 x - 28} : x < - 4 or x > 7

Extreme Points of

\frac{2 x - 5}{x ^{2} - 3 x - 28} :

Minimum

(\frac{127}{4}, \frac{6 13}{11})

Find the range for the interval

- \infty < x < - 4 : - \infty < f (x) < - 2

Find the range for the interval

7 < x < \infty : \frac{6 13}{11} \leq f (x) < \infty

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

f (x) < - 2 or f (x) \geq \frac{6 13}{11}

f (x) < - 2 or f (x) \geq \frac{6 13}{11}

f (x) = 3 \cdot e^{2 x}

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

f (x) = - 4 x^{2} - 20 x + 1

f (x) = ln (3 - 4 x^{2})

\frac{x}{x ^{2} + 1}

What is the range of (2x-5)/(sqrt(x^2-3x-28)) ?
The range of (2x-5)/(sqrt(x^2-3x-28)) is f(x)<-2\lor f(x)>= (6sqrt(13))/(11)