What is the range of f(x)=sqrt(8/x+3) ?

The range of f(x)=sqrt(8/x+3) is 0 sqrt(3)

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range of f(x)=sqrt(8/x+3)

range

f (x) = \frac{8}{x} + 3

0 \leq f (x) < 3 or f (x) > 3

Interval Notation

[0, 3) \cup (3, \infty)

Solution steps

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

Domain of

\frac{8}{x} + 3 : x \leq - \frac{8}{3} or x > 0

Extreme Points of

\frac{8}{x} + 3 :

None

Find the range for the interval

- \infty < x \leq - \frac{8}{3} : 0 \leq f (x) < 3

Find the range for the interval

0 < x < \infty : 3 < f (x) < \infty

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

0 \leq f (x) < 3 or f (x) > 3

0 \leq f (x) < 3 or f (x) > 3

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

f (x) = - x + 1 - 3

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

f (x) = \frac{7}{x} - \frac{9}{x + 9}

1.5 cos (6 x - 3.2)

What is the range of f(x)=sqrt(8/x+3) ?
The range of f(x)=sqrt(8/x+3) is 0<= f(x)<sqrt(3)\lor f(x)>sqrt(3)