What is the range of (sqrt(x-3))/(x+2) ?

The range of (sqrt(x-3))/(x+2) is 0<= f(x)<= 1/(2sqrt(5))

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range of (sqrt(x-3))/(x+2)

range

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

0 \leq f (x) \leq \frac{1}{2 5}

Interval Notation

[0, \frac{1}{2 5}]

Solution steps

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

Domain of

\frac{x - 3}{x + 2} : x \geq 3

Extreme Points of

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

Minimum

(3, 0),

Maximum

(8, \frac{1}{2 5})

Find the range for the interval

3 \leq x < \infty : 0 \leq f (x) \leq \frac{1}{2 5}

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

0 \leq f (x) \leq \frac{1}{2 5}

- x^{3} + 27 x - 54

f^{9}

f (x) = (6)^{x}

f (x) = 6 - 6 x

(\frac{1}{2})^{x - 3}

What is the range of (sqrt(x-3))/(x+2) ?
The range of (sqrt(x-3))/(x+2) is 0<= f(x)<= 1/(2sqrt(5))