What is the range of sqrt(1/x+2) ?

The range of sqrt(1/x+2) is 0 sqrt(2)

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range of sqrt(1/x+2)

range

\frac{1}{x} + 2

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

Solution steps

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

Domain of

\frac{1}{x} + 2 : x \leq - \frac{1}{2} or x > 0

Extreme Points of

\frac{1}{x} + 2 :

None

Find the range for the interval

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

Find the range for the interval

0 < x < \infty : 2 < f (x) < \infty

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

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

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

f (x) = 35 - 5 x

f (x) = \frac{2 - x ^{2}}{x ^{2} - 9}

f (x) = 8 - 3 x

x^{3} - 2 x^{2} - 15 x + 10

y + 4 = \frac{2}{3} (x - 3)

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