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

The range of 2/((sqrt(2x-5))) is f(x)>0

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range of 2/((sqrt(2x-5)))

range

\frac{2}{( 2 x - 5 )}

f (x) > 0

Interval Notation

(0, \infty)

Solution steps

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

Domain of

\frac{2}{2 x - 5} : x > \frac{5}{2}

Extreme Points of

\frac{2}{2 x - 5} :

None

Find the range for the interval

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

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

f (x) > 0

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

x^{2} - 3 x - 54

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

f (x) = 6 + 2 x - 7

y = 5 x - 2