What is the range of f(x)= 2/(sqrt(x-3)-1) ?

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

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range of f(x)= 2/(sqrt(x-3)-1)

range

f (x) = \frac{2}{x - 3 - 1}

f (x) \leq - 2 or f (x) > 0

Interval Notation

(- \infty, - 2] \cup (0, \infty)

Solution steps

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

Domain of

\frac{2}{x - 3 - 1} : 3 \leq x < 4 or x > 4

Extreme Points of

\frac{2}{x - 3 - 1} :

Maximum

(3, - 2)

Find the range for the interval

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

Find the range for the interval

4 < x < \infty : 0 < f (x) < \infty

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

f (x) \leq - 2 or f (x) > 0

f (x) \leq - 2 or f (x) > 0

y = - 3 + x

f (x) = 7 x ∣ x ∣

y = 2 x - 5

y = \frac{x ^{4}}{2}

(70.2) (80.1)

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